Question

If the radiations of wavelength 2500 A and 5000 A are incident on a substance of for function \( 2 eV \) one by one then the ratio of the stopping potentials for the emitted photoelectrons will be- nearly.

Answer 1

Given 

\(\lambda_1 = 2500\mathring A\)

\(\lambda_2= 5000\mathring A\)

\(E_ 1= hv\)

\(E_1 = \frac{hc}{\lambda_1}\)

\(E_1 = \frac{6.64 \times 10^{-34}\times 3\times 10^8}{2500\times 10^{-10}}\)

\(E_1 = \frac{19.92 \times 10^{-26}}{2500 \times 10^{-10}}\)

\(E_1 = 0.796 \times 10^{-18}\)

\(E_1 = 7.96 \times 10^{-19} V\)

\(E_1 = \frac{7.96 \times 10^{-19}}{1.6 \times 10^{-19}}\)

\(E_1 =4 .97\, lV\)

\(K_1 = E_1 - W\)

\(= 4.97 - 2\)

\(K_1 = 2.97 \,lV\)

\(lV_0 = 2.97\, lV\)

\(V_{01} = 2.97 V\)

\(E_2 = \frac{hc}{\lambda_2}\)

\(E_2 = \frac{6.64 \times 10^{-34}\times 3 \times 10^8}{5000\times 10^{-10}}\)

\(E_2 = \frac{19.92 \times 10^{-26}}{5000\times 10^{-10}}\)

\(E_2 = 3.984 \times 10^{-19}V\)

\(E_2 = 2.49 \;lV\)

\(K_2 = E_2 - W\)

\(K_2 = 2.49 - 2\)

\(K_2 = 0.49 \,lV\)

\(lV_{02} = 0.49\, lV\)

\(V_{02} = 0.49V\)

Ratio

\(\frac{V_{01}}{V_{02}} = \frac{2.97}{0.49}\)