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The peak emission from a black body at a certain temperature occurs at a wavelength of 9000Ao. On increasing its 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 in temperature, the peak radiation from the black body caused photoemission. To bring these photoelectrons to rest, potential equivalent excitation energy between n=2 and n=3 Bohr levels of hydrogen atoms is required. Find the work function of the metal.

Answer
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Hint: Here using two relationships that is energy radiated is directly proportional to temperature and there is one particular wavelength above which there is no photoelectric emission.

Complete step by step answer:
Let E1 and E2 be the initial and final rate of emission at temperature T1 and T2 respectively.
Using Stefan’s law, the total amount of heat energy E radiated per second per unit area of the surface of a perfectly black body is directly proportional to the fourth power of its absolute temperature T.
Now represented in terms of the equation,
EαT4
After removing proportionality we get, E=σT4
Where σ is called the Stefan’s constant. Its value is 5.67×108Wm2K4
Now we will consider Stefan’s equation for our further calculations.
Then, E1=σT14 and E2=σT24
From the question, after increasing the temperature heat energy emitted is 81 times the initial energy emitted at temperature T1. Then, E2=81E1
Now substitute the values of E1 and E2 in the above equation, we get
σT14=81σT24
σ gets cancel out from above equation then remaining equation is,
T14=81T24
81 can also be written as 34 then,
T14=34T24
Now taking fourth root on both sides we get,
T2=3T1
From this we can say that temperature has increased 3 times more than the initial temperature T1.
Let us find the emission from a black body at a temperature T2 occurs at a wavelength of λ2.
We have emission from a black body at a temperatureT1occurs at a wavelength of λ1=9000Ao. But at this wavelength, there is no photoelectric emission. Because we know that the wavelength of the incident radiation must be less than the threshold wavelength.
Therefore now we need to calculate λ2,
According to Wien’s displacement law, the product of wavelength (λm )at which maximum energy is emitted and absolute temperature (T) of the black body is always constant.
That is λmT=constant=b
Now consider a given problem we get
λ1T1=λ2T2
9000×T1=λ2×3T1
λ2=90003
=3000A0
At this wavelength, there is the emission of black body radiation from the surface.
Now let us calculate the work function,
Consider Einstein’s photoelectric equation,E=ϕ0+12mvmax2 ………… (1)
ϕ0 is the work function.
We have 12mvmax2=eV0
Then, equation (1) becomes,
ϕ0=EeV0
Where E is the energy of the incident radiation, that is E=hν Where ν is frequency of incident radiation is the Planck’s constant.
ϕ0=hνeV0
ϕ0=hcλ2eV0 …………… (2)
Next we will calculate the value of eV0 ,
eV0=E1E2 …………. (3)
E1=13.6n21
On substituing n value,
=13.632
On simplification, we get
 =1.51eV
Similarly,
E2=13.6n22
On substituting n value,
=13.622
On simplification, we get
 =3.4eV
Now apply these values in equation (3),
eV0=1.51(3.4)
 =1.9eV
 Now substituting all the values in equation(2) we get,ϕ0=6.625×1034×3×1083000×1010×1.6×10191.9
Here 1.6×1019 is divided because to convert energy E from joule to eV.
Therefore, the final answer is in eV.
ϕ0=4.141.9
=2.24eV

The work function is 2.24eV.

Additional information:
The minimum energy required by an electron to overcome the surface barrier of the metal is called the work function of the metal. The minimum energy required for an electron to just escape from the metal surface is called “work function”. It is denoted by (ϕ0).
The work function depends on:
(i) the nature of the metal
(ii) the conditions of the metal surface
According to Einstein’s photoelectric equation,
When a light is incident on a metal, the photons having energy hν collide with electrons at the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously. The minimum energy needed for the electron to come out of the metal surface is called work function. If the energy hν of the incident photon exceeds the work function(ϕ0) ,the electrons are emitted with a maximum kinetic energy.
hν= ϕ0+Kmax
Kmax=hνϕ0

Note:
In this type of question, the main thing is the threshold wavelength that is a certain value of incident wavelength above which there is no emission of photoelectrons. In this question, the first wavelength does not create photoelectric emission. Therefore we need to find a second wavelength.