Question

Find the ratio of the energies of two radiations of wavelengths $3000\stackrel{\circ }{A}$ and 900 nm respectively.

• A. 1:3
• B. 2:3
• C. 3:2
• D. 3:1

Answer 1

The correct option is D 3:1

According to Planck's quantum theory $E=\frac{hc}{\lambda }$
Where, E = Energy of the radiation
h = Planck's constant $=6.626\times {10}^{-34}Js$
c = speed of light $=3\times {10}^{8}m/s$
$\lambda$ = wavelength of the radiation
Given: ${\lambda }_1=3000\stackrel{\circ }{A}$ and ${\lambda }_2=900nm$ = $9000\times {10}^{-10}\stackrel{\circ }{A}$
$E_1=\frac{hc}{{\lambda }_1}....\left(i\right)$ and $E_2=\frac{hc}{{\lambda }_2}....\left(ii\right)$
Dividing equation (i) with equation (ii)
$\frac{E_1}{E_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{9000}{3000}=\frac{3}{1}$
So,
$E_1=3E_2$
Hence, the ratio of energies of the two radiations is $3:1$.
Since energy is inversely proportional to the wavelength, the one with a lower wavelength would have a higher energy.