Wave Nature of Matter

Dual Nature of Radiation and Matter · CBSE

Theory

What this section has to deliver

de Broglie's argument in words, the relation $\lambda = h/p$ with its accelerated-particle forms, the consistency check with a photon, and the ratio problems. Two- and three-mark questions here are nearly always a ratio of de Broglie wavelengths for proton, deuteron, alpha particle or electron.

Duality, in both directions

The wave nature of light shows up in interference, diffraction and polarisation. The particle nature shows up in the photoelectric effect and the Compton effect, which involve energy and momentum transfer.
Which description fits depends on the experiment. Seeing an object needs both: the eye lens gathers and focuses light, which the wave picture describes; the rods and cones absorb it, which needs the photon picture.
de Broglie's argument (1924), to write out in the exam: nature is symmetrical, and matter and radiation are the two basic physical entities of the universe, so they should be symmetrical in character. If radiation has a dual wave-particle nature, matter should have one too. Moving particles of matter should therefore display wave-like properties under suitable conditions.

The de Broglie relation

$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$

$\lambda$ = de Broglie wavelength (m), $p$ = momentum (kg m s $^{-1}$ ), $m$ = mass (kg), $v$ = speed (m s $^{-1}$ ). The waves associated with moving material particles are called matter waves or de Broglie waves.
The duality is contained in the equation itself: $\lambda$ on the left is a wave attribute, $p$ on the right a particle attribute, and Planck's constant relates the two.
The de Broglie wavelength is independent of the charge and of the nature of the particle. It depends only on the momentum.
It is measurable only for sub-atomic particles, whose masses and momenta are small enough to give wavelengths of the order of atomic spacings in crystals. For everyday objects it is hopelessly small.

Consistency with the photon

For a photon, $p = h\nu/c$ . Substituting into the de Broglie relation:

$\lambda = \dfrac{h}{p} = \dfrac{h}{h\nu/c} = \dfrac{c}{\nu}$

which is exactly the wavelength of the electromagnetic radiation of which the photon is a quantum. The de Broglie relation, invented for matter, reproduces the right answer for radiation.

Forms for accelerated and energetic particles

$\lambda = \dfrac{h}{\sqrt{2mK}} \qquad \lambda = \dfrac{h}{\sqrt{2mqV}} \qquad p = \sqrt{2mK} = \sqrt{2mqV}$

$K$ = kinetic energy in joule; $q$ = charge in coulomb; $V$ = accelerating potential difference in volt, using $K = qV$ for a particle starting from rest. Non-relativistic only.
For an electron this reduces to the number worth memorising:

$\lambda = \dfrac{12.27}{\sqrt{V}}\ \text{\text{Å}} = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$

At $V = 100$ V, $\lambda = 1.227$ Å. At $V = 50$ kV, $\lambda \approx 5.49\times10^{-12}$ m.
Dependences to state when asked: $\lambda \propto 1/v$ , $\lambda \propto 1/\sqrt K$ , $\lambda \propto 1/\sqrt V$ , $\lambda \propto 1/p$ . Increasing the accelerating potential decreases $\lambda$ ; decreasing the speed increases $\lambda$ .

Graphs

$p$ against $\lambda$ : from $p = h/\lambda$ , a rectangular hyperbola — a decreasing curve, concave upwards, approaching but never touching either axis. Never a straight line.
$\lambda$ against kinetic energy: from $\lambda = h/\sqrt{2mK}$ , so $\lambda^2 K$ is constant. A steeply falling curve that flattens out as $K$ grows, never touching either axis.
$\lambda$ against accelerating potential: same shape, since $\lambda \propto V^{-1/2}$ .

Worked ratio problems

Comparison	Condition	Result
Proton and $\alpha$ -particle	same accelerating potential	$\lambda_p : \lambda_\alpha = 2\sqrt2 : 1$
Deuteron and $\alpha$ -particle	same accelerating potential	$\lambda_d : \lambda_\alpha = 2 : 1$
Proton and $\alpha$ -particle	same kinetic energy	$\lambda_p : \lambda_\alpha = 2 : 1$
Proton and $\alpha$ -particle	same speed	$\lambda_p : \lambda_\alpha = 4 : 1$
Proton and deuteron	same speed	$\lambda_p : \lambda_d = 2 : 1$
Proton and electron	same speed	$\lambda_p : \lambda_e = 1 : 1836$
Proton and $\alpha$ -particle	same $\lambda$	$V_p : V_\alpha = 8 : 1$ and $v_p : v_\alpha = 4 : 1$
Two fragments of a body at rest	momentum conservation	$\lambda_1 : \lambda_2 = 1 : 1$

Method for the first row: $\lambda = h/\sqrt{2mqV}$ , so at the same $V$ , $\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\dfrac{4m_p\times2e}{m_p\times e}} = \sqrt8 = 2\sqrt2$ .
Method for the second row: a deuteron has $m_d = 2m_p$ and charge $e$ , an alpha particle $m_\alpha = 4m_p = 2m_d$ and charge $2e$ , so $\dfrac{\lambda_d}{\lambda_\alpha} = \sqrt{\dfrac{2m_d\times2e}{m_d\times e}} = \sqrt4 = 2$ .
Method for the last row: a body at rest splitting in two gives $m_1v_1 = -m_2v_2$ , so the momenta are equal in magnitude, so $\lambda_1 = \lambda_2$ whatever the masses.
Accelerated through the same potential, $\lambda \propto 1/\sqrt m$ and $p \propto \sqrt m$ : of a proton and a deuteron, the proton has the greater wavelength and the smaller momentum. Of an electron and a proton, the electron has the greater wavelength and the smaller momentum.

Other standard results

Photon and electron of the same wavelength $\lambda$ : the electron's kinetic energy is $E_e = \dfrac12 mv^2 = \dfrac{h^2}{2m\lambda^2}$ and the photon's energy is $E_p = \dfrac{hc}{\lambda}$ , so

$\dfrac{E_p}{E_e} = \dfrac{hc}{\lambda}\times\dfrac{2m\lambda^2}{h^2} = \dfrac{2\lambda mc}{h} \qquad \Longrightarrow \qquad E_p = \dfrac{2\lambda mc}{h}E_e$

Photoemission by X-rays with the work function negligible: all the photon energy becomes kinetic energy, $E_\nu = \dfrac12 mv^2$ , so $mv = \sqrt{2mE_\nu}$ and the emitted electron's de Broglie wavelength is $\lambda = \dfrac{h}{\sqrt{2mE_\nu}}$ , i.e. $\lambda \propto E_\nu^{-1/2}$ . The graph of $\lambda$ against $E_\nu$ is the same steeply falling hyperbola-like curve.
Electron microscope: at $50$ kV the electron wavelength is about $5.49\times10^{-12}$ m, while yellow light is about $5700$ to $5900$ Å, i.e. of order $10^{-7}$ m — some $10^{5}$ times larger. Resolving power varies inversely with the wavelength used, so the electron microscope's resolving power is about $10^{5}$ times that of an optical microscope, other factors such as numerical aperture being the same.

Why matter waves are invisible in daily life

A ball of mass $0.12$ kg at $20$ m s $^{-1}$ : $p = 2.40$ kg m s $^{-1}$ and $\lambda = 2.76\times10^{-34}$ m — beyond any measurement.
A ball of mass $0.150$ kg at $30.0$ m s $^{-1}$ : $p = 4.50$ kg m s $^{-1}$ and $\lambda = 1.47\times10^{-34}$ m, about $10^{-19}$ times the size of a proton.
An electron at $5.4\times10^{6}$ m s $^{-1}$ : $p = 4.92\times10^{-24}$ kg m s $^{-1}$ and $\lambda = 0.135$ nm — comparable to X-ray wavelengths, and to the spacing of atomic planes in a crystal, so it is measurable.
The lesson: $\lambda$ is smaller for a heavier or faster particle, which is why the wave character of matter is significant only in the sub-atomic domain.

A caution NCERT adds

The wavelength of a matter wave, $\lambda = h/p$ , has physical significance. The phase velocity of the matter wave has none. The group velocity is physically meaningful and equals the velocity of the particle.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
How will the de Broglie wavelength associated with an electron be affected when (i) the velocity of the electron decreases, and (ii) the accelerating potential is increased? Justify your answer.
1 markCBSE Class 12 Physics — Very Short Answer · 2023 · Section A, 1 markAnswer 1 ↓
2
An electron is accelerated through a potential difference of 100 V. Calculate the de Broglie wavelength associated with it.
1 markCBSE Class 12 Physics — Very Short Answer · 2021 · Section A, 1 markAnswer 2 ↓
3
A proton and a deuteron are moving with the same speed. Find the ratio of the de Broglie wavelength ( $\lambda_p/\lambda_d$ ) associated with them.
1 markCBSE Class 12 Physics — Very Short Answer · 2021 · Section A, 1 markAnswer 3 ↓
4
A proton and an electron have equal speeds. Find the ratio of de Broglie wavelengths associated with them.
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 4 ↓
5
Draw a plot showing the variation of de-Broglie wavelength of an electron as a function of its K.E.
1 markCBSE Class 12 Physics — Very Short Answer · Delhi 2015C, U · Section A, 1 markAnswer 5 ↓
6
What are matter waves? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the de Broglie wavelength associated with the proton to that with the alpha particle.
2 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section B, 2 marksAnswer 6 ↓
7
(a) Calculate the energy and momentum of a photon in a monochromatic beam of wavelength 331.5 nm. (b) How fast should a hydrogen atom travel in order to have the same momentum as that of the photon in part (a)?
2 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section B, 2 marksAnswer 7 ↓
8
An $\alpha$ -particle and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths.
2 marksCBSE Class 12 Physics — SA · Delhi 2017 · Section B, 2 marksAnswer 8 ↓
9
The wavelength $\lambda$ of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of a photon is $(2\lambda mc/h)$ times the kinetic energy of the electron, where $m$ , $c$ and $h$ have their usual meaning.
2 marksCBSE Class 12 Physics — SA · Foreign 2016, An · Section B, 2 marksAnswer 9 ↓
10
A proton and an $\alpha$ -particle have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating potentials, (ii) their speeds.
2 marksCBSE Class 12 Physics — SA · Delhi 2015 · Section B, 2 marksAnswer 10 ↓
11
A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has (a) greater value of de-Broglie wavelength associated with it, and (b) less momentum? Give reasons to justify your answer.
2 marksCBSE Class 12 Physics — SA · Delhi 2014 · Section B, 2 marksAnswer 11 ↓
12
X-rays fall on a photosensitive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de-Broglie wavelength ( $\lambda$ ) of the electrons emitted and the energy ( $E_\nu$ ) of the incident photons. Draw the nature of the graph for $\lambda$ as a function of $E_\nu$ .
2 marksCBSE Class 12 Physics — SA · Delhi 2014C · Section B, 2 marksAnswer 12 ↓
13
(a) A proton and an $\alpha$ -particle are moving with the same speed. Which one of them has the greater value of de Broglie wavelength associated with it? Justify your answer. (b) Lights of wavelengths 430 nm, 450 nm and 660 nm are incident on a metal surface whose threshold wavelength is 600 nm. In which case/cases will photoemission take place, and why? Calculate the threshold frequency of the metal surface.
3 marksCBSE Class 12 Physics — SA · 2022C · Section C, 3 marksAnswer 13 ↓
14
Obtain an expression for the ratio of the accelerating potentials required to accelerate a proton and an $\alpha$ -particle to have the same de-Broglie wavelength associated with them.
3 marksCBSE Class 12 Physics — SA · AI 2019 · Section C, 3 marksAnswer 14 ↓
15
(a) An electron and a proton are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it, and (ii) lesser momentum? Justify your answer in each case. (b) How is the momentum of a particle related with its de-Broglie wavelength? Show the variation on a graph.
3 marksCBSE Class 12 Physics — SA · AI 2019 · Section C, 3 marksAnswer 15 ↓
16
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
3 marksCBSE Class 12 Physics — SA · AI 2014 · Section C, 3 marksAnswer 16 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

17
A photon and an alpha particle have the same kinetic energy. The ratio of de Broglie wavelengths associated with the proton to that with the alpha particle is
- (a)
  1
- (b)
  2
- (c)
  $2\sqrt2$
- (d)
  $\dfrac12$
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 17 ↓
18
Which of the following graphs correctly represents the variation of a particle's momentum with its associated de-Broglie wavelength?
- (a)
  $p$ vs $\lambda$ : straight line through origin, positive slope
- (b)
  $p$ vs $\lambda$ : concave curve rising steeply near $\lambda=0$
- (c)
  $p$ vs $\lambda$ : straight line, negative slope
- (d)
  $p$ vs $\lambda$ : rectangular hyperbola (decreasing curve, concave up)
1 markCBSE Class 12 Physics — Multiple Choice · 2023, U · Section A, 1 markAnswer 18 ↓
19
A heavy particle initially at rest splits spontaneously into two particles of masses $m_1$ and $m_2$ having non-zero velocities. The ratio of de Broglie wavelengths associated with the particles is
- (a)
  $m_1/m_2$
- (b)
  $m_2/m_1$
- (c)
  1
- (d)
  $\sqrt{m_2}/\sqrt{m_1}$
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 19 ↓
20
The graph showing the correct variation of linear momentum ( $p$ ) of a charged particle with its de-Broglie wavelength ( $\lambda$ ) is
- (a)
  $p$ vs $\lambda$ : straight line, negative slope through the axes (linear decrease)
- (b)
  $p$ vs $\lambda$ : rectangular-hyperbola-like decreasing curve, concave up
- (c)
  $p$ vs $\lambda$ : increasing concave curve starting from origin
- (d)
  $p$ vs $\lambda$ : straight line through origin, positive slope
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 20 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
De Broglie wavelength: $\;\lambda=\dfrac h{mv}$ (i) When velocity decreases, $\lambda$ INCREASES (since $\lambda\propto1/v$ ). (ii) $\;\lambda=\dfrac h{mv}=\dfrac h{\sqrt{2mK}}=\dfrac h{\sqrt{2meV_0}}\;\Rightarrow\;\lambda\propto\dfrac1{\sqrt{V_0}}$ . When the accelerating potential increases, the de Broglie wavelength DECREASES.
2.↑ back to the question
$\;\lambda=\dfrac{12.27}{\sqrt V}\text{ \text{Å}}=\dfrac{12.27}{\sqrt{100}}\text{ \text{Å}}=1.227\text{ \text{Å}}$
3.↑ back to the question
$\;\lambda=\dfrac h{mv_0}$ ; with $m_d=2m_p$ : $\;\lambda_p=\dfrac h{m_pv_0}$ , $\;\lambda_d=\dfrac h{2m_pv_0}$ $\;\dfrac{\lambda_p}{\lambda_d}=\dfrac{h/(m_pv_0)}{h/(2m_pv_0)}=2\;\Rightarrow\;\lambda_p:\lambda_d=2:1$
4.↑ back to the question
$\;\lambda=\dfrac h{mv}\;\Rightarrow\;\dfrac{\lambda_p}{\lambda_e}=\dfrac{m_e}{m_p}=\dfrac1{1836}$
5.↑ back to the question
$\;\lambda=\dfrac hp=\dfrac h{\sqrt{2mKE}}\;\Rightarrow\;\lambda\propto\dfrac1{\sqrt{KE}}$ , so $\lambda^2\times KE=$ constant. The graph of $\lambda$ vs $KE$ is a rectangular-hyperbola-like decreasing curve (steeply falling as $KE$ increases from zero, then flattening out), never touching either axis.
6.↑ back to the question
According to Louis de Broglie, a wave is associated with every moving particle; these are called matter waves. de Broglie wavelength of a charged particle accelerated through a potential difference $V$ : $\;\lambda=\dfrac h{\sqrt{2mqV}}$ For proton and $\alpha$ -particle charges are $q$ and $2q$ respectively, and $m_\alpha=4m_p$ : $\;\dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt{\dfrac{4m_p\times2q}{m_p\times q}}=\sqrt8=2\sqrt2$
7.↑ back to the question
(a) Energy: $\;E=\dfrac{hc}\lambda=\dfrac{1240\text{ nm-eV}}{331.5\text{ nm}}\approx3.74\text{ eV}$ Momentum: $\;p=\dfrac Ec=\dfrac{3.74\times1.6\times10^{-19}}{3\times10^8}\approx2\times10^{-27}\text{ kg m/s}$ (b) For a hydrogen atom of mass $m\approx1.6\times10^{-27}$ kg: $\;v=\dfrac pm=\dfrac{2\times10^{-27}}{1.6\times10^{-27}}=1.24\text{ m/s}$
8.↑ back to the question
$\;\lambda=\dfrac h{\sqrt{2mqV}}$ ; for proton ( $m,q$ ) and $\alpha$ -particle ( $4m,2q$ ): $\;\dfrac{\lambda_\alpha}{\lambda_p}=\sqrt{\dfrac{mq}{4m\times2q}}=\sqrt{\dfrac18}=\dfrac1{2\sqrt2}$ , i.e. $\;\lambda_p:\lambda_\alpha=2\sqrt2:1$ .
9.↑ back to the question
de Broglie wavelength of electron: $\;\lambda=\dfrac h{mv}\;\Rightarrow\;v=\dfrac h{m\lambda}$ Kinetic energy of electron: $\;E_e=\dfrac12mv^2=\dfrac12m\left(\dfrac h{m\lambda}\right)^2=\dfrac{h^2}{2m\lambda^2}\quad\dots(i)$ Energy of photon of the same wavelength $\lambda$ : $\;E_p=\dfrac{hc}\lambda\quad\dots(ii)$ Dividing (ii) by (i): $\;\dfrac{E_p}{E_e}=\dfrac{hc}\lambda\times\dfrac{2m\lambda^2}{h^2}=\dfrac{2m\lambda c}h$ $\therefore\;E_p=\dfrac{2\lambda mc}hE_e$ , as required.
10.↑ back to the question
$\;\lambda=\dfrac h{\sqrt{2mqV}}$ . Here $m_p=m$ , $q_p=e$ , $m_\alpha=4m$ , $q_\alpha=2e$ . (i) Since $\lambda_p=\lambda_\alpha$ : $\;m_pq_pV_p=m_\alpha q_\alpha V_\alpha\;\Rightarrow\;m\times e\times V_p=4m\times2e\times V_\alpha\;\Rightarrow\;V_p=8V_\alpha\;\Rightarrow\;V_p:V_\alpha=8:1$ (ii) $\;\lambda=\dfrac h{mv}\;\Rightarrow\;v=\dfrac h{m\lambda}$ ; for the same $\lambda$ : $\;\dfrac{v_p}{v_\alpha}=\dfrac{m_\alpha}{m_p}=\dfrac{4m}m=4\;\Rightarrow\;v_p:v_\alpha=4:1$
11.↑ back to the question
$\;\lambda=\dfrac h{\sqrt{2mqV}}\propto\dfrac1{\sqrt m}$ (for the same $q,V$ , assuming charge $q=e$ for both). (a) Since the deuteron's mass is greater than the proton's, the proton (smaller mass) has the GREATER de Broglie wavelength. (b) Momentum $p=\sqrt{2mqV}\propto\sqrt m$ . Since the deuteron's mass is greater, the proton (smaller mass) has LESS momentum.
12.↑ back to the question
Neglecting the work function, by Einstein's photoelectric equation: $\;E_\nu=\dfrac12mv^2\;\Rightarrow\;mv=\sqrt{2mE_\nu}$ de Broglie wavelength of the emitted electron: $\;\lambda=\dfrac h{mv}=\dfrac h{\sqrt{2mE_\nu}}$ , i.e. $\;\lambda\propto\dfrac1{\sqrt{E_\nu}}$ . The graph of $\lambda$ vs $E_\nu$ is a steeply decreasing curve (rectangular-hyperbola-like, $\lambda^2E_\nu=$ constant), never touching either axis.
13.↑ back to the question
(a) $\;\lambda=\dfrac h{mv}\propto\dfrac1m$ (for the same $v$ ). With $m_\alpha=4m_p$ : $\;\lambda_p=\dfrac h{m_pv}$ , $\;\lambda_\alpha=\dfrac h{4m_pv}$ , so $\lambda_p>\lambda_\alpha$ ; the PROTON, having less mass, has the greater de Broglie wavelength. (b) Energies of the incident photons using $E=\dfrac{1240\text{ eV nm}}\lambda$ : for $\lambda=430$ nm, $E_1=2.88$ eV; for $\lambda=450$ nm, $E_2=2.76$ eV; for $\lambda=660$ nm, $E_3=1.88$ eV. Threshold energy for $\lambda_0=600$ nm: $\;E_0=\dfrac{1240}{600}\approx2.07$ eV. Since $E_1(2.88\text{ eV})$ and $E_2(2.76\text{ eV})$ both exceed $E_0$ , photoemission takes place for wavelengths 430 nm and 450 nm; since $E_3(1.88\text{ eV})<E_0$ , photoemission does NOT occur for 660 nm. Threshold frequency: $\;\nu_0=\dfrac c{\lambda_0}=\dfrac{3\times10^8}{600\times10^{-9}}=5\times10^{14}\text{ Hz}$
14.↑ back to the question
$\;\lambda=\dfrac h{\sqrt{2mqV}}$ . Let proton ( $m_p,q_p=e$ ) and $\alpha$ -particle ( $m_\alpha=4m_p,q_\alpha=2e$ ) have the same $\lambda$ . $\;\dfrac h{\sqrt{2m_pq_pV_p}}=\dfrac h{\sqrt{2m_\alpha q_\alpha V_\alpha}}\;\Rightarrow\;m_pq_pV_p=m_\alpha q_\alpha V_\alpha$ $\;m_p\times e\times V_p=4m_p\times2e\times V_\alpha\;\Rightarrow\;V_p=8V_\alpha\;\Rightarrow\;\dfrac{V_p}{V_\alpha}=8:1$
15.↑ back to the question
(a) de Broglie wavelength for a particle of charge $q$ accelerated through potential $V$ : $\;\lambda=\dfrac h{\sqrt{2mK}}=\dfrac h{\sqrt{2m(qV)}}$ , so $\lambda\propto1/\sqrt m$ (for the same $q,V$ ). (i) As the electron's mass is much less than the proton's, the ELECTRON has the greater de Broglie wavelength. (ii) Momentum $p=\sqrt{2mK}\propto\sqrt m$ . As the electron's mass is lesser than the proton's, the momentum of the ELECTRON is lesser than that of the proton. (b) $\;\lambda=\dfrac hp\;\Rightarrow\;p=\dfrac h\lambda$ , i.e. $p\propto1/\lambda$ — momentum is inversely proportional to de Broglie wavelength. It shows a rectangular hyperbola in the $p$ - $\lambda$ graph.
16.↑ back to the question
$\;\lambda=\dfrac h{\sqrt{2meV}}=\dfrac{12.27}{\sqrt V}\text{ \text{Å}}$ $\;\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50\times10^3}}\approx5.49\times10^{-12}\text{ m}$ For a yellow-light optical microscope, wavelength $\approx5700$ – $5900$ Å, i.e. $\sim10^{-7}$ m — roughly $10^5$ times larger than the electron's de Broglie wavelength found above ( $\sim10^{-12}$ m). Resolving power of a microscope is INVERSELY proportional to the wavelength used. Since the electron's de Broglie wavelength is vastly smaller than that of yellow light, the resolving power of an electron microscope is MUCH GREATER than that of an optical microscope using yellow light.
17.Answer: (B)↑ back to the question
De Broglie wavelength: $\;\lambda=\dfrac hp$ , and $p=\sqrt{2mE}\;\Rightarrow\;\lambda=\dfrac h{\sqrt{2mE}}$ For the same kinetic energy $E$ : $\;\lambda\propto\dfrac1{\sqrt m}\;\Rightarrow\;\dfrac{\lambda_{proton}}{\lambda_{alpha}}=\sqrt{\dfrac{m_{alpha}}{m_{proton}}}$ Since $m_{alpha}=4m_{proton}$ : $\;\dfrac{\lambda_{proton}}{\lambda_{alpha}}=\sqrt4=2$ . Option (b).
18.Answer: (D)↑ back to the question
$\;\lambda=\dfrac hp\;\Rightarrow\;p\propto\dfrac1\lambda$ : the graph is a rectangular hyperbola. Option (d).
19.Answer: (C)↑ back to the question
By conservation of momentum, since the particle was initially at rest: $\;0=m_1v_1+m_2v_2\;\Rightarrow\;m_1v_1=-m_2v_2\;\Rightarrow\;|p_1|=|p_2|$ $\;\lambda=\dfrac hp\;\Rightarrow\;\dfrac{\lambda_1}{\lambda_2}=\dfrac{p_2}{p_1}=1$ . Option (c).
20.Answer: (B)↑ back to the question
$\;p=\dfrac h\lambda\;\Rightarrow\;p\propto\dfrac1\lambda$ , i.e. momentum is inversely proportional to wavelength — the graph is a rectangular hyperbola (decreasing, concave up). Option (b).