MHT CET 2020 16th October Morning Shift | Dual Nature of Radiation Question 82 | Physics

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

$$\frac{v^2}{2\left(\frac{e}{m}\right)}$$

$$\frac{v^2}{\left(\frac{m}{e}\right)}$$

$$\frac{v^2}{2\left(\frac{m}{e}\right)}$$

$$\frac{v^2}{\left(\frac{e}{m}\right)}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-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

$$1: 4$$

$$1: 2$$

$$1: \sqrt{2}$$

$$1: 1$$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

The stopping potential of the photoelectrons, from a photo cell is

directly proportional to intensity of incident light

directly proportional to frequency of incident light

inversely proportional to frequency of incident light

Inversely proportional to intensity of incident light

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

When certain metal surface is illuminated with a light of wavelength $\lambda$, the stopping potential is $V$, When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is $\left(\frac{V}{3}\right)$. The threshold wavelength for the surface is

$\frac{8 \lambda}{3}$

$\frac{4 \lambda}{3}$

$4 \lambda$

$6 \lambda$

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