Question

The energy of a photon is given as $\Delta E/atom=3.03\times {10}^{-19}ȷ{atom}^{-1}$. Then the wavelength $\left(\lambda \right)$ of the photon is:

• A. 65.6 nm
• B. 656 nm
• C. 0.656 nm
• D. 6.56 nm

Answer 1

Answer: (B)

656 nm

Energy of photon of $=3.03\times {10}^{-19}j$

$\frac{hc}{\lambda }=3.03\times {10}^{-19}$

$\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{\lambda }=3.03\times {10}^{-19}$

$\lambda =\frac{6.626\times 3\times {10}^{-26}}{3.03\times {10}^{-19}}$

$6.56\times {10}^{-7}$

$=656nm$

Ans $B$