1
COMEDK 2024 Morning Shift
+1
-0

The mass of a particle $$\mathrm{A}$$ is double that of the particle $$\mathrm{B}$$ and the kinetic energy of $$\mathrm{B}$$ is $$\frac{1}{8}$$th that of A then the ratio of the de- Broglie wavelength of A to that of B is:

A
1 : 2
B
2 : 1
C
1 : 4
D
4 : 1
2
COMEDK 2024 Morning Shift
+1
-0

The difference in energy levels of an electron at two excited levels is $$13.75 \mathrm{~eV}$$. If it makes a transition from the higher energy level to the lower energy level then what will be the wave length of the emitted radiation? [given $$h=6.6 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} \mathrm{~s}^{-1} ; c=3 \times 10^8 \mathrm{~ms}^{-1} ; 1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{~J}$$]

A
$$900 \mathrm{~nm}$$
B
$$9^0 \mathrm{~A}$$
C
$$9000 \mathrm{~nm}$$
D
$$900^{\circ} \mathrm{A}$$
3
COMEDK 2023 Morning Shift
+1
-0

When a certain metal surface is illuminated with light of frequency $$\nu$$, the stopping potential for photoelectric current is $$V_0$$. When the same surface is illuminated by light of frequency $$\frac{\nu}{2}$$, the stopping potential is $$\frac{V_0}{4}$$. The threshold frequency for photoelectric emission is

A
$$\frac{\nu}{6}$$
B
$$\frac{\nu}{3}$$
C
$$\frac{2\nu}{3}$$
D
$$\frac{4\nu}{3}$$
4
COMEDK 2023 Morning Shift
+1
-0

Let $$K_1$$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $$\lambda_1$$ and $$K_2$$ corresponding to wavelength $$\lambda_2$$. If $$\lambda_1=2 \lambda_2$$, then

A
$$2 K_1=K_2$$
B
$$K_1=2 K_2$$
C
$$K_1 < K_2 / 2$$
D
$$K_1>2 K_2$$
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