1
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+1
-0

The maximum velocity of the photoelectron emitted by the metal surface is $$v$$. Charge and mass of the photoelectron is denoted by $$e$$ and $$m$$, respectively. The stopping potential in volt is

A
$$\frac{v^2}{2\left(\frac{e}{m}\right)}$$
B
$$\frac{v^2}{\left(\frac{m}{e}\right)}$$
C
$$\frac{v^2}{2\left(\frac{m}{e}\right)}$$
D
$$\frac{v^2}{\left(\frac{e}{m}\right)}$$
2
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+1
-0

Energy of the incident photon on the metal surface is $$3 W$$ and then $$5 W$$, where $$W$$ is the work function for that metal. The ratio of velocities of emitted photoelectrons is

A
$$1: 4$$
B
$$1: 2$$
C
$$1: \sqrt{2}$$
D
$$1: 1$$
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