Question

If the energy of a photon corresponding to a wavelength of $6000\,\mathring{A}$ is $3.32 \times 10^{-19} \,J$, the photon energy for a wavelength of $4000 \,\mathring{A}$ will be:

• A. 1.4 eV
• B. 4.9 eV
• C. 3.1 eV
• D. 1.6 eV

Answer 1

Answer: (C)

The energy of a photon is given as
$E = hv =\frac{h c}{\lambda}$
Given, $\lambda_{1}=6000 \,\mathring{A}$
$E_{1}=3.32 \times 10^{-19} J$
$\lambda_{2}=4000 \,\mathring{A}$
$\therefore \frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$
$\Rightarrow \frac{3.32 \times 10^{-19}}{E_{2}}=\frac{4000}{6000}$
$\Rightarrow E_{2}=\frac{6}{4} \times 3.32 \times 10^{-19} J$
$=4.98 \times 10^{-19} J$
$=\frac{4.98 \times 10^{-19}}{1.6 \times 10^{-19}} eV$
$=3.1\, eV$