Question

The wavelength (in $\,\mathring{A}$) of a photon having energy $3 \,eV$ is approximately.
$\left[1\, eV =1.6 \times 10^{-12} erg \right]$
$\left[h=6.626 \times 10^{-27} erg\, s \right]$

• A. 3000
• B. 4000
• C. 4141
• D. 7824

Answer 1

Answer: (C)

$3\, eV =3 \times 1.6 \times 10^{-12}$ erg $=$ Energy $(E)$

$\lambda=\frac{h c}{E}=\frac{6.626 \times 10^{-27} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{-12}}=414 1$ ( $\left. \,\mathring{A}\right)$