Question

The energy of a photon of wavelength $6600\mathring{A}$ in watt-hour is :-

• A. $3\times 10^{-19}$
• B. $1.875$
• C. $8.33\times 10^{-23}$
• D. $5\times 10^{-21}$

Answer 1

Answer: (C)

$E=\frac{hc}{\lambda }=\frac{12400}{6600}eV$
$=\frac{12400}{6600}\times 1.6\times 10^{-19}J$
$\left(1eV=1.6\times 10^{-19}J\right)$
$\Rightarrow E=\frac{12400}{6600}\times 1.6\times 10^{-19}$ watt-sec
$(1J=1$ watt-sec $)$
$\Rightarrow E=\frac{12400}{6600}\times 1.6\times 10^{-19}\frac{W\text{ att }-\text{ hour }}{3600}$
$[1$ watt $-\sec =\frac{1}{3600}.$ watt $-$ hour $]$
$\Rightarrow E=8.33\times 10^{-23}$