Question

wavelength associated with a photon having energy 4.14 eV is

• A. $3000A^{\circ }$
• B. $4000A^{\circ }$
• C. $5000A^{\circ }$
• D. $5800A^{\circ }$

Answer 1

Answer: (A)

$\lambda =\frac{hc}{E}=\frac{6.62\times 10^{-34}\times 3\times 10^8}{4.14\times 1.6\times 10^{-19}}m=3000A^{\circ }$ or $\lambda =\frac{12400}{4.14}{A}^{\circ }$ nearly