Question

An electron in a stationary Bohr orbit of hydrogen atom jumps from 4th energy level to the ground state. The velocity that the photon acquired as a result of electron transition will be (h = Planck's constant, R = Rydberg's constant, m = mass of photon)

• A. \( \frac{15 h R}{16 m} \)
• B. \( \frac{12 h R}{13 m} \)
• C. \( \frac{9 h R}{11 m} \)
• D. \( \frac{7 h R}{9 m} \)

Hint:

The energy difference in the Bohr model for an electron transition between two orbits can be calculated using the Rydberg formula.

Answer 1

The Correct Option is A

Step 1: Use the energy difference in the Bohr model.
The energy difference between the 4th and the ground state (n=1) for a hydrogen atom is given by the formula: \[ E = -R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For the 4th energy level (\( n_1 = 4 \)) and the ground state (\( n_2 = 1 \)), the energy difference is: \[ E = -R \left( 1 - \frac{1}{16} \right) = -R \times \frac{15}{16} \] Step 2: Relating energy to photon velocity.
The energy of a photon is related to its velocity by: \[ E = \frac{h \nu}{2 \pi} = \frac{h c}{\lambda} \] where \( c \) is the speed of light and \( \lambda \) is the wavelength of the emitted photon. By equating the energy expressions, we can find the velocity of the photon, which will be proportional to \( \frac{15 h R}{16 m} \). Step 3: Conclusion.
Thus, the velocity that the photon acquired as a result of the electron transition is \( \frac{15 h R}{16 m} \), corresponding to option (A).