Question

If an electron in the excited state falls to ground state, a photon of energy 5 eV is emitted, then the wavelength of the photon is nearly

• A. 748 nm
• B. 598 nm
• C. 398 nm
• D. 248 nm

Hint:

Use \(E = \frac{hc}{\lambda}\) and convert units properly to find wavelength from photon energy.

Answer 1

The Correct Option is D

Step 1: Energy-wavelength relation
Energy \(E\) of photon and wavelength \(\lambda\) are related by: \[ E = \frac{hc}{\lambda} \] where \(h = 6.63 \times 10^{-34} J \cdot s\), \(c = 3 \times 10^8 m/s\).
Step 2: Convert 5 eV to Joules
\[ 1 eV = 1.6 \times 10^{-19} J \implies 5 eV = 8 \times 10^{-19} J \] Step 3: Calculate wavelength
\[ \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{8 \times 10^{-19}} = 2.48 \times 10^{-7} m = 248 nm \] Step 4: Conclusion
The wavelength is approximately 248 nm.