MHT CET 2024 2nd May Morning Shift | Dual Nature of Radiation Question 58 | Physics

When a metallic surface is illuminated with a radiation of wavelength ' $\lambda$ ', the stopping potential is ' $V$ '. If the same surface is illuminated with radiation of wavelength ' $3 \lambda$ ', the stopping potential is ' $\left(\frac{\mathrm{V}}{6}\right)$ '. The threshold wavelength for the surface is

$3 \lambda$

$4 \lambda$

$5 \lambda$

$6 \lambda$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

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The work function of metal ' $A$ ' and ' $B$ ' are in the ratio $1: 2$. If light of frequency ' $f$ ' and ' $2 f$ ' is incident on surface ' $A$ ' and ' $B$ ' respectively, then the ratio of kinetic energies of emitted photo electrons is

$1: 1$

$1: 2$

$1: 3$

$1: 4$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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When radiation of wavelength '$$\lambda$$' is incident on a metallic surface, the stopping potential is 4.8 V. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes $$1.6 \mathrm{~V}$$. The threshold wavelength for the surface is

$$2 \lambda$$

$$4 \lambda$$

$$6 \lambda$$

$$8 \lambda$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The de-Broglie wavelength $$(\lambda)$$ of a particle is related to its kinetic energy (E) as

$$\lambda \propto \mathrm{E}$$

$$\lambda \propto \mathrm{E}^{-1}$$

$$\lambda \propto \mathrm{E}^{\frac{1}{2}}$$

$$\lambda \propto \mathrm{E}^{-\frac{1}{2}}$$

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