MHT CET 2020 16th October Evening Shift | Dual Nature of Radiation Question 88 | 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 47 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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