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)

$E =\frac{ hc }{\lambda}=\frac{12400}{6600} eV$
$=\frac{12400}{6600} \times 1.6 \times 10^{-19} J$
$\left(1\, eV =1.6 \times 10^{-19} J \right)$
$\Rightarrow E =\frac{12400}{6600} \times 1.6 \times 10^{-19}$ watt-sec
$(1\, J =1$ 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 $]$
$\Rightarrow E =8.33 \times 10^{-23}$