question_answer 71)

Following results were observed when sodium metal is irradiated with different wavelengths. Calculate: (a) Threshold wavelength (b) Planck's constant \[\begin{align} & \lambda nm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,500\,\,\,\,\,\,\,\,450\,\,\,\,\,\,\,\,\,\,\,400 \\ & \upsilon \times {{10}^{6}}m{{s}^{-1}}\,\,\,\,\,2.55\,\,\,\,\,\,4.35\,\,\,\,\,\,\,\,\,5.20 \\ \end{align}\]  

Answer:

We know that: Absorbed Threshold Kinetic energy of -          = energy energy photoelectrons \[hc\left[ \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right]=\frac{1}{2}m{{v}^{2}}\] \[\frac{hc}{{{10}^{-9}}}\left[ \frac{1}{500}-\frac{1}{{{\lambda }_{0}}} \right]=\frac{1}{2}m{{(2.55\times {{10}^{6}})}^{2}}\,\,\,\,\,\,\,\,\,\,......(i)\] \[\frac{hc}{{{10}^{-9}}}\left[ \frac{1}{450}-\frac{1}{{{\lambda }_{0}}} \right]=\frac{1}{2}m{{(4.35\times {{10}^{6}})}^{2}}\,\,\,\,\,\,\,\,\,\,......(ii)\] \[\frac{hc}{{{10}^{-9}}}\left[ \frac{1}{400}-\frac{1}{{{\lambda }_{0}}} \right]=\frac{1}{2}m{{(5.20\times {{10}^{6}})}^{2}}\,\,\,\,\,\,\,\,\,\,......(iii)\]Dividing Eqn. (ii) by Eqn. (i), we get \[\left[ \frac{{{\lambda }_{0}}-450}{{{\lambda }_{0}}\times 450} \right]\times \left[ \frac{500-{{\lambda }_{0}}}{{{\lambda }_{0}}-500} \right]={{\left( \frac{4.35}{2.55} \right)}^{2}}\] On solving, we get; \[{{\lambda }_{0}}=531nm\] Putting the value of \[{{\lambda }_{0}}\] in equation (iii), we get; \[\frac{h\times 3\times {{10}^{8}}}{{{10}^{-9}}}\left[ \frac{1}{400}-\frac{1}{531} \right]=\frac{1}{2}\times {{9.10}^{-31}}{{(5.20\times {{10}^{6}})}^{2}}\] \[h=6.65\times {{10}^{-32}}Js\]