Vardaan Institute - Physics Solutions Key (Level 1)

Topic 11.1: Photoelectric Effect & Einstein's Equation

1.Answer:

$\Phi_0 = h\nu_0 \implies \nu_0 = \Phi_0 / h$.
$\Phi_0 = 2.14 \times 1.6 \times 10^{-19} \text{ J}$.
$\nu_0 = \frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx \mathbf{5.16 \times 10^{14} \text{ Hz}}$.

2.Answer:

$eV_0 = h(\nu - \nu_0) = h\Delta\nu$.
$\Delta\nu = (8.2 - 3.3) \times 10^{14} = 4.9 \times 10^{14} \text{ Hz}$.
$V_0 = \frac{6.63 \times 10^{-34} \times 4.9 \times 10^{14}}{1.6 \times 10^{-19}} \approx \mathbf{2.03 \text{ V}}$.

3.Answer:

In wave theory, energy depends on intensity. Any frequency of light, given enough intensity and time, should eventually transfer enough energy to an electron to eject it. Wave theory cannot explain why emission stops completely below a specific frequency regardless of intensity.

4.Answer:

Energy of incident light $E = \frac{hc}{\lambda} = \frac{1242 \text{ eV}\cdot\text{nm}}{488 \text{ nm}} \approx 2.545 \text{ eV}$.
$K_{max} = eV_0 = 0.38 \text{ eV}$.
$\Phi_0 = E - K_{max} = 2.545 - 0.38 = \mathbf{2.165 \text{ eV}}$.

5.Answer:

The slope represents the ratio $\mathbf{h/e}$ (Planck's constant divided by electron charge), which is universal for all metals.

6.Answer:

$E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19} \text{ J}$.
$E(\text{eV}) = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx \mathbf{2.49 \text{ eV}}$.

7.Answer:

It **remains unchanged**. Stopping potential depends only on the frequency of the light and the nature of the material. Changing the distance only changes the intensity (number of photons), not the energy of individual photons.

8.Answer:

(i) $K_{max}$ remains unchanged as it depends only on frequency.
(ii) Photoelectric current increases linearly as more photons strike the surface per second.

9.Answer:

Energy of photon $E = \frac{1242}{330} \approx 3.76 \text{ eV}$.
Since $E < \Phi_0$ ($3.76 < 4.2$), **no emission** will occur.

10.Answer:

$P = n E_{photon} \implies n = P / h\nu$.
$n = \frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \approx \mathbf{5 \times 10^{15} \text{ photons/s}}$.

11.Answer:

1. Instantaneous emission.
2. Existence of threshold frequency.
3. Independence of $K_{max}$ from intensity.

12.Answer:

Stopping potential is the minimum negative voltage required to stop even the fastest photoelectron. **No**, it does not depend on intensity.

13.Answer:

$h\nu = 6.63 \times 10^{-34} \times 10^{15} = 6.63 \times 10^{-19} \text{ J} \approx 4.14 \text{ eV}$.
$K_{max} = 4.14 - 2.5 = 1.64 \text{ eV} = 2.624 \times 10^{-19} \text{ J}$.
$v = \sqrt{2K_{max}/m} = \sqrt{2 \times 2.624 \times 10^{-19} / 9.1 \times 10^{-31}} \approx \mathbf{7.6 \times 10^5 \text{ m/s}}$.

14.Answer:

This happens when the energy of the incident photons is only slightly higher than the work function. Electrons are emitted with very little kinetic energy and cannot reach the collector against the small retarding field produced by the contact potential unless a positive bias is applied.

15.Answer:

It represents the **threshold frequency ($\nu_0$)** for that specific material.

Topic 11.2: Matter Waves (de Broglie Hypothesis)

16.Answer:

$\lambda = \frac{12.27}{\sqrt{100}} = \mathbf{1.227 \text{ \AA}}$ (or $0.123 \text{ nm}$).

17.Answer:

$\lambda \propto 1/\sqrt{m}$. The alpha particle has the highest mass, so it has the **shortest wavelength**.

18.Answer:

$\lambda_p / \lambda_e = (m_e v_e) / (m_p v_p) \implies m_p = m_e \times (v_e / v_p) \times (\lambda_e / \lambda_p)$.
$m_p = m_e \times (1/3) \times (1 / 1.813 \times 10^{-4}) \approx 1836 m_e$.
The particle is a **proton**.

19.Answer:

$\lambda = \frac{12.27}{\sqrt{100}} = \mathbf{1.227 \text{ \AA}}$.

20.Answer:

(Image Solution)

AI Image Prompt: A clean, high-quality, mathematically correct landscape graph showing the variation of de Broglie wavelength (lambda) with the magnitude of momentum (p). The graph must show a rectangular hyperbola (lambda inversely proportional to p) in the first quadrant. Label the axes clearly. The background of the whole image should be fully white.

Filename: Level1_Q20_Lambda_vs_Momentum_Graph.png

21.Answer:

$\lambda = \frac{6.63 \times 10^{-34}}{0.15 \times 30} \approx \mathbf{1.47 \times 10^{-34} \text{ m}}$.
This wavelength is so small (much smaller than the size of a nucleus) that it cannot be observed with any physical aperture or crystal.

22.Answer:

$K = p^2 / 2m \implies p = \sqrt{2mK}$.
From de Broglie's hypothesis: $\lambda = h/p$.
Substituting $p$: $\lambda = \frac{h}{\sqrt{2mK}}$.

23.Answer:

$\lambda = h / \sqrt{2mK} \implies K = h^2 / (2m\lambda^2)$.
For same $\lambda$, $K \propto 1/m$. Therefore, particle **B** (lower mass) has higher kinetic energy.

24.Answer:

$1 = 12.27 / \sqrt{V} \implies \sqrt{V} = 12.27 \implies V \approx \mathbf{150.5 \text{ V}}$.

25.Answer:

$K = \frac{3}{2} k_B T$. $\lambda = \frac{h}{\sqrt{2m(1.5 k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.
Substituting values ($T=300\text{K}$): $\lambda \approx \mathbf{1.45 \text{ \AA}}$.

26.Answer:

Photon: Rest mass is zero, always moves at speed $c$.
Electron: Finite rest mass, velocity is always less than $c$.

27.Answer:

$\lambda \propto 1 / \sqrt{m}$.
$\lambda_e / \lambda_p = \sqrt{m_p / m_e} \approx \sqrt{1836} \approx \mathbf{42.8}$.

28.Answer:

Since $\lambda = h/p$, if $p$ is doubled, the wavelength is **halved**.

29.Answer:

Matter waves represent the probability of finding a particle. Unlike EM waves, they don't involve oscillating fields, and unlike acoustic waves, they don't require a medium or bulk pressure changes. They exist only for moving particles.

30.Answer:

Photon energy $E_p = hc/\lambda$. Electron energy $E_e = p^2/2m = h^2/2m\lambda^2$.
For $\lambda \sim 1 \text{ \AA}$, $E_p \approx 12.4 \text{ keV}$ while $E_e \approx 150 \text{ eV}$. The **photon** usually has significantly higher energy.

Topic 11.3: Davisson-Germer Experiment

31.Answer:

Objective: To confirm the wave nature of electrons. Conclusion: Electrons exhibit diffraction patterns just like X-rays, proving matter waves exist.

32.Answer:

A single crystal provides large, well-defined atomic planes in a specific orientation, which is necessary to produce a distinct and sharp diffraction peak at a predictable angle.

33.Answer:

Accelerating Voltage: **$54 \text{ V}$**. Scattering Angle: **$50^\circ$**.

34.Answer:

$\theta = 90 - \phi/2 = 90 - 25 = \mathbf{65^\circ}$.

35.Answer:

(Image Solution)

AI Image Prompt: A clean, high-quality, mathematically correct landscape polar plot showing electron scattering intensity versus scattering angle (phi) for four different voltages: 40V, 44V, 48V, and 54V. The 54V plot should show a very distinct, large 'bump' at 50 degrees, while the others show smaller or no bumps. The background of the whole image should be fully white.

Filename: Level1_Q35_Davisson_Germer_Voltage_Comparison.png

36.Answer:

Bragg's: $\lambda = 2d \sin\theta = 2(0.91) \sin 65^\circ \approx 1.82 \times 0.906 \approx \mathbf{1.65 \text{ \AA}}$.
de Broglie: $\lambda = 12.27 / \sqrt{54} \approx \mathbf{1.67 \text{ \AA}}$.
The two values match within experimental error.

37.Answer:

Because the wavelength derived from the purely wave-based Bragg diffraction formula matched the wavelength predicted by de Broglie's particle-based formula for the same set of electrons.

38.Answer:

The intensity increases significantly, developing a pronounced "bump" that reaches its absolute maximum at $54 \text{ V}$.

39.Answer:

Electron Gun: Produces a collimated, mono-energetic beam of electrons. Faraday Cylinder: A collector that can rotate to measure the current of electrons scattered at different angles.

40.Answer:

**No**. Protons are $\sim 1836$ times heavier. To achieve the same wavelength (needed for diffraction at $50^\circ$ on the same crystal), the voltage required would be much lower ($V \propto 1/m$).

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