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Q.

A peak emission from a black body at a certain temperature occurs at a wavelength of 9000A°. On increasing the temperature, the total radiation emitted is increased 81 times. At the initial temperature when the peak radiation from the black body is incident on a metal surface, it does not cause any photoemission from the surface. After the increase of temperature the peak radiation from the black body caused photoemission. To bring these photoelectrons to rest, a potential equivalent to the excitation energy between then n=2 and n=3 Bohr levels of hydrogen is required. Find the work function of the metal.

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a

3.78eV

b

3.27eV

c

2.24eV

d

2.76eV

answer is A.

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Detailed Solution

Let T be the initial temperature of the black body. The total energy emitted by the body per unit area per second is given by

E=σT4

Where σ is the Stefan’s constant. When the temperature of the black body is raised to T', we have

E'=σT'4

Now, E'E=81  (given).

Hence, 81=T'T4

Which gives T'=3T

It is given that peak emission at temperature T occurs at a wavelength λm=9000A°. If λ'm is the wavelength for peak emission at temperature T', then from Wien’s displacement law λmT=constant, we have 

λ'mT'=λmT

or λ'm=λmTT'=9000A°×T3T=3000A°

According to the problem, the kinetic energy of the emitted photo-electrons = excitation energy for transition n=2 to n=3, which is

KEmax=E23=13.6eV122-132=13.6×536=1.89eV

The energy of the photon of wavelength λ'=3000A° in eV is

E=124003000eV=4.13eV

Now, from Einstein’s photoelectric equation, the work function for the metal surface is given as 

ϕ=E-KEmax=4.13-1.89eV   =2.24eV

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