MHT CET 2023 9th May Morning Shift | Dual Nature of Radiation Question 20 | Physics

From a metallic surface photoelectric emission is observed for frequencies $$v_1$$ and $$v_2\left(v_1 > v_2\right)$$ of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $$1: \mathrm{x}$$. Hence the threshold frequency of the metallic surface is

$$\frac{v_1-v_2}{x}$$

$$\frac{v_1-v_2}{x-1}$$

$$\frac{x v_1-v_2}{x-1}$$

$$\frac{x v_2-v_1}{x-1}$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

-0

If the kinetic energy of a free electron doubles, it's de Broglie wavelength ($$\lambda$$) changes by a factor

$$\frac{1}{\sqrt{2}}$$

$$\frac{1}{2}$$

$$\sqrt{2}$$

$$2$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

When a photon enters glass from air, which one of the following quantity does not change?

Energy

Velocity

Wavelength

Momentum

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

$$\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$$

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