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
$E=\frac{h c}{\lambda}$
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 $
$1\, eV =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 $
$\Rightarrow E=1.89 \,eV$