The de - Broglie wavelength associated with the electron in the $n = 4$ level is :

Question

The de-Broglie wavelength of a moving bus with speed $ v $ is $ \lambda $. Some passengers left the bus at a stoppage. Now when the bus moves with twice of its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

Answer 1

To determine the de-Broglie wavelength of an electron at a specific energy level, we can use the principles of quantum mechanics and the de-Broglie hypothesis. According to the de-Broglie hypothesis, the wavelength (\lambda) associated with a particle is given by:

\lambda = \frac{h}{p}

where h is Planck's constant and p is the momentum of the particle.

For an electron in a hydrogen-like atom, the momentum is related to the principal quantum number (n). The momentum (p) of the electron in the n-th orbit is given by:

p = \frac{m ev_n}{n}

where m is the mass of the electron, e is the charge of the electron, and v_n is the velocity of the electron in the n-th orbit.

The velocity (v_n) can be expressed as:

v_n = \frac{e^2}{2 \varepsilon_0 h n}

where \varepsilon_0 is the permittivity of free space.

Substituting this back into the expression for momentum, the de-Broglie wavelength in the n-th orbit becomes:

\lambda_n = \frac{h}{m \cdot \frac{m e \cdot v_n}{n}} = \frac{2 \pi \varepsilon_0 h^2 n^2}{m e^2}

This implies that the de-Broglie wavelength is proportional to n^2.

Therefore, for the n = 4 level:

\lambda_4 = \lambda_1 \times \left(\frac{4^2}{1^2}\right)

\lambda_4 = 16 \lambda_1

This shows that the de-Broglie wavelength in the n = 4 level is four times that in the ground state, justifying the correct answer as:

four times the de-Broglie wavelength of the electron in the ground state

Answer 2