Question

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

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

Answer 1

Answer: (A)

(a) $\because E=h v=\frac{h c}{\lambda}=h c \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) $\rightarrow E=6.6 \times 10^{-19} J$
For (b) $E=h v=6.63 \times 10^{-34} \times 3 \times 10^{8}$
(b) $\rightarrow E=1.98 \times 10^{-25} J$
For (c) $E=h c \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) $d \rightarrow E=6.62 \times 10^{-27} J$
Hence, highest energy for photon is in $(a)$.