Question

The difference in energy levels of an electron at two excited levels is 13.75 eV. If it makes a transition from the higher energy level to the lower energy level then what will be the wavelength of the emitted radiation? 
Given:
$ h = 6.6 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{s}^{-1} $, $ c = 3 \times 10^8 \, \text{ms}^{-1} $, $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $

• A. 900 nm
• B. \( 90^\circ \, \text{A} \)
• C. 9000 nm
• D. 900  $\text{A}$

Hint:

To calculate the wavelength of emitted radiation, use the formula \( \lambda = \frac{h \cdot c}{E} \), and make sure to convert the energy from eV to Joules when necessary.

Answer 1

The Correct Option is A

The energy \( E \) of a photon is given by the formula: \[ E = \frac{h \cdot c}{\lambda} \] Where: - \( E \) is the energy of the photon, - \( h \) is Planck's constant, - \( c \) is the speed of light, and - \( \lambda \) is the wavelength. Rearranging the formula for wavelength \( \lambda \): \[ \lambda = \frac{h \cdot c}{E} \] Substituting the given values: - The energy difference \( E = 13.75 \, \text{eV} = 13.75 \times 1.6 \times 10^{-19} \, \text{J} \), - \( h = 6.6 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{s}^{-1} \), - \( c = 3 \times 10^8 \, \text{ms}^{-1} \). 
Now, calculating the wavelength: \[ \lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{13.75 \times 1.6 \times 10^{-19}} \] \[ \lambda = 9.0 \times 10^{-7} \, \text{m} = 900 \, \text{nm} \] Thus, the wavelength of the emitted radiation is 900 nm.