Question

The peak emission from a black body at a certain temperature occurs at a wavelength of $9000\mathring{A}$. 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 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 $n=2$ and $n=3$ Bohr levels of hydrogen atoms is required. Find the work function of the metal.

Answer 1

Let $E_1$ and $E_2$ be the rates of emission at temperatures $T_1$ and $T_2$ respectively.

Using Stefan's law, $E_1=\sigma T_1^4$ and $E_2=\sigma T_2^4$

Given, $E_2=81E_1\Rightarrow T_2^4=81T_1^4$ or $T_2=3T_1$

Using Wien's law , ${\lambda }_1{T}_1={\lambda }_2{T}_2$

or $9000T_1={\lambda }_2\left(3T_1\right)$

or ${\lambda }_2=3000A^o$

From photoelectric effect, $e{V}_0=\frac{hc}{{\lambda }_2}-W$

or $W=\frac{hc}{{\lambda }_2}-e{V}_0$

Here, $e{V}_0=-13.6(1/3^2-1/2^2)=1.89eV$ and $\frac{hc}{{\lambda }_2}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-10}}J=6.62\times {10}^{-19}J=\frac{6.62\times {10}^{-19}}{1.6\times {10}^{-19}}=4.13eV$

Thus, $W=4.13-1.89=2.25eV$