JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Modern Physics

WhatsApp Email

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is

A \times 1 0^{Z}

(where

1 < A < 10

). The value of ' Z ' is

Answer: 7

Solution

Energy of incident photon:

E = \frac{h c}{λ} = \frac{1240}{200} = 6.2

eV. Maximum kinetic energy of photoelectrons:

K E_{ma x} = 6.2 - 4.7 = 1.5

eV. This becomes the stopping potential. For a sphere of radius r = 1 cm:

V = \frac{k q}{r} = 1.5

V. Number of electrons:

n = \frac{V r}{k e} = \frac{1.5 \times 0.01}{9 \times 1 0 ^{9} \times 1.6 \times 1 0 ^{- 19}} \approx 1.05 \times 1 0^{7}

. Therefore Z = 7.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer

Practise more: all JEE Advanced PYQs on Modern Physics →

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Modern Physics

Solution

More JEE Advanced questions on Modern Physics