Question

A quantum of light having energy E has wavelength equal to $7200A^0$. The frequency of light which corresponds to energy equal to 3E is:

• A. $1.25\times {10}^{14}{s}^{-1}$
• B. $1.25\times {10}^{15}{s}^{-1}$
• C. $1.25\times {10}^{13}{s}^{-1}$
• D. $1.25\times {10}^{16}{s}^{-1}$

Answer 1

Answer: (B)

$1.25\times {10}^{15}{s}^{-1}$

$E\propto \frac{1}{wavelength}\propto frequency$
$wavelength=\frac{1}{3}\left(7200\right)A^0$
$frequency=\frac{C}{\lambda }=\frac{3\times 3\times {10}^{8}m/s}{7200\times {10}^{-10}m}$
$=1.25\times {10}^{15}{s}^{-1}$