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;
\[\frac{h\times 3\times {{10}^{8}}}{{{10}^{-9}}}\left[
\frac{1}{400}-\frac{1}{531} \right]=\frac{1}{2}\times
{{9.10}^{-31}}{{(5.20\times {{10}^{6}})}^{2}}\] \[h=6.65\times
{{10}^{-32}}Js\]
You need to login to perform this action.
You will be redirected in
3 sec