Light of wavelength $496 n m$ is incident on a metal surface causing ejection of photoelectrons for slopping potential $1.5 V$ , then

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The de-Broglie wavelength of fastest photoelectron $100 n m$

to move the fastest electron in a circle of radius $1 m$ , perpendicular magnetic field required $4 μ T$

to move the fastest electron in a circle of radius im , perpendicular magnetic field required 8 $μ T$

the work function of the surface is $1 e V$

answer is A, C.

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Detailed Solution

$K_{\max} = 1.5 e V λ = \frac{h}{\sqrt{2 m K}} \Rightarrow λ = 1 n m E of incident photon = \frac{1241}{496} = 2.5 e V \Rightarrow ϕ = 2.5 - 1.5 = 1 e V Since, q v B = \frac{m v^{2}}{r} \Rightarrow B = \frac{m v_{m a x}}{q r} = \frac{\sqrt{2 m K_{m a x}}}{q r} \Rightarrow B = \frac{\sqrt{2 \times 9.1 \times 10^{- 31} \times e \times 1.5}}{e \times 1} = 4.13 \times 10^{- 6} T = 4 μ T Hence, (1) and (3) are correct.$