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Let ${E_1}$ and ${E_2}$ be the initial and final rate of emission at temperature ${T_1}$ and ${T_2}$ 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,

$E\alpha {T^4}$

After removing proportionality we get, $E = \sigma {T^4}$

Where $\sigma $ is called the Stefan’s constant. Its value is $5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$

Now we will consider Stefan’s equation for our further calculations.

Then, ${E_1} = \sigma {T_1}^4$ and ${E_2} = \sigma {T_2}^4$

From the question, after increasing the temperature heat energy emitted is 81 times the initial energy emitted at temperature ${T_1}$. Then, ${E_2} = 81{E_1}$

Now substitute the values of ${E_1}$ and ${E_2}$ in the above equation, we get

$\sigma {T_1}^4 = 81\sigma {T_2}^4$

$\sigma $ gets cancel out from above equation then remaining equation is,

${T_1}^4 = 81{T_2}^4$

81 can also be written as ${3^4}$ then,

${T_1}^4 = {3^4}{T_2}^4$

Now taking fourth root on both sides we get,

${T_2} = 3{T_1}$

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 ${\lambda _2}$.

We have emission from a black body at a temperatureT1occurs at a wavelength of ${\lambda _1} = 9000\mathop A\limits^o $. 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 ${\lambda _2}$,

According to Wien’s displacement law, the product of wavelength (${\lambda _m}$ )at which maximum energy is emitted and absolute temperature (T) of the black body is always constant.

That is ${\lambda _m}T = $constant=$b$

Now consider a given problem we get

${\lambda _1}{T_1} = {\lambda _2}{T_2}$

$9000 \times {T_1} = {\lambda _2} \times 3{T_1}$

${\lambda _2} = \dfrac{{9000}}{3}$

$ = 3000\mathop {\text{A}}\limits^{\text{0}} $

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,$E = {\phi _0} + \dfrac{1}{2}m{v_{\max }}^2$ ………… (1)

${\phi _0}$ is the work function.

We have $\dfrac{1}{2}m{v_{\max }}^2 = e{V_0}$

Then, equation (1) becomes,

${\phi _0} = E - e{V_0}$

Where E is the energy of the incident radiation, that is $E = h\nu $ Where $\nu $ is frequency of incident radiation is the Planck’s constant.

${\phi _0} = h\nu - e{V_0}$

${\phi _0} = h\dfrac{c}{{{\lambda _2}}} - e{V_0}$ …………… (2)

Next we will calculate the value of $e{V_0}$ ,

$e{V_0} = {E_1} - {E_2}$ …………. (3)

${E_1} = \dfrac{{ - 13.6}}{{{n^2}_1}}$

On substituing $n$ value,

$ = \dfrac{{ - 13.6}}{{{3^2}}}$

On simplification, we get

$ = - 1.51eV$

Similarly,

${E_2} = \dfrac{{ - 13.6}}{{{n^2}_2}}$

On substituting $n$ value,

$ = \dfrac{{ - 13.6}}{{{2^2}}}$

On simplification, we get

$ = - 3.4eV$

Now apply these values in equation (3),

$e{V_0} = - 1.51 - ( - 3.4)$

$ = 1.9eV$

Now substituting all the values in equation(2) we get,${\phi _0} = 6.625 \times {10^{ - 34}} \times \dfrac{{3 \times {{10}^8}}}{{3000 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}} - 1.9$

Here $1.6 \times {10^{ - 19}}$ is divided because to convert energy E from joule to eV.

Therefore, the final answer is in eV.

${\phi _0} = 4.14 - 1.9$

$ = 2.24eV$

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 $\left( {{\phi _0}} \right)$.

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 $h\nu$ 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 $h\nu$ of the incident photon exceeds the work function\[\left( {{\phi _0}} \right)\] ,the electrons are emitted with a maximum kinetic energy.

\[h\nu = {\text{ }}{\phi _0} + {K_{\max }}\]

\[{K_{\max }} = h\nu - {\phi _0}\]

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.

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