Question

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is n × 10^-4 eV. The value of n is____.[Assume the atom to be stationary during the absorption and emission process](Take h = 6.6 x 10^-34 js and c = 3 x 10⁸ m/s)

Answer 1

Correct Answer: 4125

The net energy absorbed by the atom is given by: \[ E = E_i - E_e = h c \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right), \] where: \( h = 6.6 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), \( \lambda_1 = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m} \), \( \lambda_2 = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \). Substitute the values: \[ E = \left(6.6 \times 10^{-34}\right) \cdot \left(3 \times 10^8\right) \cdot \left(\frac{1}{5 \times 10^{-7}} - \frac{1}{6 \times 10^{-7}}\right). \] Simplify: \[ E = \left(6.6 \times 10^{-34} \cdot 3 \times 10^8\right) \cdot \left(\frac{6 - 5}{30 \times 10^{-7}}\right). \] \[ E = \left(1.98 \times 10^{-25}\right) \cdot \left(\frac{1}{5 \times 10^{-7}}\right). \] Convert to eV (\( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)): \[ E = \frac{1.98 \times 10^{-25}}{1.6 \times 10^{-19}} = 4.125 \times 10^{-4} \, \text{eV}. \] Thus, the net energy absorbed is \( \boxed{4.125 \times 10^{-4} \, \text{eV}} \) and the value of \( n \) is \( \boxed{4125} \).