Question

Which one of the following corresponds to a photon of highest energy?

• A. $\lambda =300nm$
• B. $v=3\times 10^8{s}^{-1}$
• C. $\bar{v}=30c{m}^{-1}$
• D. $E=6.626\times 10^{-27}J$

Answer 1

Answer: (A)

(a) $\because E=hv=\frac{hc}{\lambda }=hc\cdot \bar{v}$
$\left(\because \frac{1}{\lambda }=\bar{v}\right)$
where $E=$ energy of photon
$c=$ velocity of photon $(=$ light $)$
$\lambda =$ wavelength of photon
$h=$ plank's constant.
$\therefore$ For (a)
$E=6.63\times 10^{-34}\times 3\times 10^8/300\times 10^{-9}$
$E=1.98\times 10^{-25}\cdot J/300\times 10^{-9}$
$\therefore E=\frac{1.98\times 10^{-25}J}{300\times 10^{-9}m}$
(a) $\to E=6.6\times 10^{-19}J$
For (b) $E=hv=6.63\times 10^{-34}\times 3\times 10^8$
(b) $\to E=1.98\times 10^{-25}J$
For (c) $E=hc\times \bar{v}$
$\left(\because \frac{1}{\lambda }=\bar{v}\right)$
$E=6.63\times 10^{-34}\times 3\times 10^8\times 30\times 10^{-2}$
(c) $\Rightarrow E=5.96\times 10^{-26}$
For (d) $d\to E=6.62\times 10^{-27}J$
Hence, highest energy for photon is in $(a)$.