Question

The energy associated with electron in first orbit of hydrogen atom is \(-2.18 \times 10^{-18}\) J. The frequency of the light required (in Hz) to excite the electron to fifth orbit is (\(h=6.6 \times 10^{-34}\) Js)

• A. \(2.17 \times 10^{16}\)
• B. \(3.17 \times 10^{14}\)
• C. \(2.17 \times 10^{15}\)
• D. \(3.17 \times 10^{15}\)

Hint:

When calculating the energy difference for absorption (excitation), the result \(\Delta E\) must be positive. You can ensure this by using \(\Delta E = |E_1|(\frac{1}{n_i^2} - \frac{1}{n_f^2})\). This avoids sign errors.

Answer 1

The Correct Option is C

Step 1: Calculate Energy of 5th Orbit: Energy of \( n \)-th orbit \( E_n = \frac{E_1}{n^2} \). \( E_1 = -2.18 \times 10^{-18} \) J. \( E_5 = \frac{-2.18 \times 10^{-18}}{5^2} = \frac{-2.18 \times 10^{-18}}{25} \) J.
Step 2: Calculate Energy Difference (\( \Delta E \)): To excite from \( n=1 \) to \( n=5 \): \[ \Delta E = E_5 - E_1 = -2.18 \times 10^{-18} \left( \frac{1}{25} - 1 \right) \] \[ \Delta E = 2.18 \times 10^{-18} \left( 1 - \frac{1}{25} \right) = 2.18 \times 10^{-18} \times \frac{24}{25} \] \[ \Delta E = 2.18 \times 0.96 \times 10^{-18} \approx 2.09 \times 10^{-18} \text{ J} \]
Step 3: Calculate Frequency (\( \nu \)): \[ \Delta E = h\nu \implies \nu = \frac{\Delta E}{h} \] \[ \nu = \frac{2.09 \times 10^{-18}}{6.6 \times 10^{-34}} = 0.317 \times 10^{16} \text{ Hz} \] \[ \nu = 3.17 \times 10^{15} \text{ Hz} \]