1.
Answer: (b) Photoelectric effect
2.
Answer: (b) Work function
3.
Answer: (b) Electron volt (eV)
4.
Answer: Threshold
5.
Answer: Intensity
6.
Answer: False (It depends on the frequency of incident light, not intensity).
7.
Answer: True
8.
Answer: Photon (or Quantum)
9.
Answer: $E = h\nu$
10.
Answer: $K_{max} = h\nu - \Phi_0$
11.
Answer: (b) Stopping potential
12.
Answer: (b) $e V_0$
13.
Answer: (Image Solution)
AI Image Prompt: A clean, high-quality, mathematically correct landscape graph showing the variation of Photoelectric Current (y-axis) with Anode Potential (x-axis) for two different INTENSITIES of incident light (I1 and I2, where I2 > I1) but at the SAME FREQUENCY. The graph must show a common negative Stopping Potential (-V0) on the x-axis, branching out into two parallel saturation current levels. The background of the whole image should be fully white.
Filename: Level0_Q13_Photoelectric_Current_vs_Potential.png
14.
Answer: $\text{J}\cdot\text{s}$ (Joule-seconds)
15.
Answer: Stopping (or Retarding)
16.
Answer: True
17.
Answer: False (It demonstrates the particle/photon nature of light).
18.
Answer: A -> 3; B -> 1; C -> 2; D -> 4
19.
Answer: Alkali
20.
Answer: $\Phi_0 = \frac{hc}{\lambda_0}$
21.
Answer: (c) Louis de Broglie
22.
Answer: (a) $\lambda = \frac{h}{mv}$
23.
Answer: (b) $p = \sqrt{2mK}$
24.
Answer: Matter (or de Broglie)
25.
Answer: Inversely
26.
Answer: False (They are not produced by accelerating charges and do not require a medium, unlike em waves; they are probability waves).
27.
Answer: False (Macroscopic objects have a $\lambda$ so incredibly small that it is unobservable).
28.
Answer: $eV$
29.
Answer: $\lambda = \frac{h}{\sqrt{2mK}}$
30.
Answer: (b) $\frac{12.27}{\sqrt{V}} \text{ \AA}$
31.
Answer: (Image Solution)
AI Image Prompt: A clean, high-quality, mathematically correct landscape graph plotting the de Broglie wavelength (lambda) on the y-axis against the term (1/sqrt(V)) on the x-axis for an accelerated electron. The graph must strictly show a straight line passing through the origin, demonstrating direct proportionality. The background of the whole image should be fully white.
Filename: Level0_Q31_DeBroglie_Wavelength_Graph.png
32.
Answer: Decreases
33.
Answer: $\frac{h}{\lambda}$
34.
Answer: False (Since $m_p > m_e$, and $\lambda \propto 1/\sqrt{m}$, the proton has a shorter wavelength).
35.
Answer: True
36.
Answer: A -> 3; B -> 1; C -> 2; D -> 4
37.
Answer: Dual
38.
Answer: Momentum
39.
Answer: $p = \frac{h\nu}{c}$ (or $E/c$)
40.
Answer: $K = \frac{p^2}{2m}$
41.
Answer: (Image Solution)
AI Image Prompt: A clean, high-quality, mathematically correct landscape schematic diagram of the Davisson-Germer experimental setup. It must clearly show an Electron Gun emitting a beam, a Nickel (Ni) target crystal, and a movable circular Electron Detector (Faraday Cylinder) collecting scattered electrons. Include angular labels for the scattering angle. The background of the whole image should be fully white.
Filename: Level0_Q41_Davisson_Germer_Setup.png
42.
Answer: (b) Wave nature of electrons
43.
Answer: (c) Nickel
44.
Answer: (b) $54 \text{ V}$
45.
Answer: $50$ (or $50^\circ$)
46.
Answer: Interference (or Diffraction)
47.
Answer: True
48.
Answer: True (Both yield approximately $1.65 - 1.67 \text{ \AA}$).
49.
Answer: Cylinder
50.
Answer: $n\lambda = 2d \sin\theta$ (or $\lambda = 2d \sin\theta$ for $n=1$)
51.
Answer: (Image Solution)
AI Image Prompt: A clean, high-quality, mathematically correct landscape polar graph of the Davisson-Germer experimental results. It must show a pronounced 'bump' or lobe (intensity peak) at exactly a 50-degree scattering angle. Label the accelerating voltage explicitly as 54V. The background of the whole image should be fully white.
Filename: Level0_Q51_Davisson_Germer_Polar_Graph.png
52.
Answer: Tungsten
53.
Answer: $1.67$
54.
Answer: False (The L.T. battery only heats the filament; a High Tension (H.T.) battery provides the massive accelerating voltage).
55.
Answer: True
56.
Answer: A -> 2; B -> 1; C -> 4; D -> 3
57.
Answer: Scattering
58.
Answer: Current (or Intensity)
59.
Answer: $\theta = 90^\circ - \frac{\phi}{2}$
60.
Answer: Vacuum
