Question

The energy in eV of red light of wavelength $\lambda =6560\mathring{A}$ is:

• A. 1.89 eV
• B. 2.89 eV
• C. 3.89 eV
• D. 4.89 Ev

Answer 1

Answer: (A)

From Planck's law, the energy $(E)$ of a wave of wavelength $\lambda$ is
$E=\frac{hc}{\lambda }$
where, $h$ is Planck's constant,
$c$ is speed of light,
and $\lambda$ is wavelength.
Given, $h=6.625\times 10^{-34}J-s$
$c=3\times 10^{8}m/s,$
$\lambda =6560\mathring{A}=6560\times 10^{-10}m$
$1eV=1.6\times 10^{-19}J$
$\therefore E=\frac{6.625\times 10^{-34}\times 3\times 10^8}{6560\times 10^{-10}\times 1.6\times 10^{-19}}eV$
$\Rightarrow E=1.89eV$