MHT CET 2020 16th October Evening Shift | Dual Nature of Radiation Question 45 | Physics

The light of wavelength $$\lambda$$ incident on the surface of metal having work function $$\phi$$ emits the electrons. The maximum velocity of electrons emitted is [ $$c=$$ velocity of light, $$h=$$ Planck's constant, $$m=$$ mass of electron]

$$\left[\frac{2(h v-\phi) \lambda}{m c}\right]$$

$$\left[\frac{2(h c-\lambda \phi)}{m \lambda}\right]^{1 / 2}$$

$$\left[\frac{2(h c-\lambda)}{m \lambda}\right]^{1 / 2}$$

$$\left[\frac{2(h c-\phi)}{m \lambda}\right]$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

The graph of kinetic energy against the frequency $$v$$ of incident light is as shown in the figure. The slope of the graph and intercept on $$X$$-axis respectively are

MHT CET 2020 16th October Evening Shift Physics - Dual Nature of Radiation Question 1 English

maximum KE threshold frequency

Planck's constant, threshold frequency

Planck's constant, work function

work function, maximum KE

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-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

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

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

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

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

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

$$1: 4$$

$$1: 2$$

$$1: \sqrt{2}$$

$$1: 1$$

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