A) \[F=\frac{hc}{d_{{}}^{2}}\]
B) \[F=\frac{hc}{d_{{}}^{4}}\]
C) \[F=\frac{hd_{{}}^{2}}{c}\]
D) \[F=\frac{d_{{}}^{4}}{hc}\]
Correct Answer: B
Solution :
[b] \[F=\frac{Newton}{{{m}^{2}}}\] \[\frac{hc}{{{d}^{2}}}=joule-\sec .\left( \frac{meter}{\sec .} \right)\frac{1}{{{(metre)}^{2}}}=\frac{joul}{metre}=Newton\]\[\frac{hc}{{{d}^{4}}}=\left( \frac{hc}{{{d}^{2}}} \right)\frac{1}{{{d}^{2}}}=\frac{Newton}{{{m}^{2}}}\]. Hence F=\[hc/{{d}^{4}}\] \[\frac{hd}{c}=\frac{Joul.\sec {{(metre)}^{2}}}{\left( \frac{metre}{\sec .} \right)}=Newton-metr{{e}^{2}}{{\sec }^{2}}\] \[\frac{{{d}^{4}}}{hc}=\frac{1}{\left( \frac{hc}{{{d}^{4}}} \right)}=\frac{{{m}^{2}}}{Newton}\]
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