Sankalp NEET Full Test-3 Question-34 Solution

Question: 34. De-Broglie wavelength of an electron orbiting in the n=2 state of hydrogen atom is close to (Given Bohr radius =0.052 \mathrm{~nm} )

(1) 0.067 nm

(2) 0.67 nm

(3) 1.67 nm

(4) 2.67 nm

Answer: Option (2)

Explanation:

According to Bohr’s model and de Broglie’s hypothesis,

the electron behaves like a standing wave around the nucleus.

Hence, the circumference of the orbit must be an integral multiple of the de Broglie wavelength.

This condition is given by:

2 \pi r_n = n \lambda

where r_n is the radius of the n^{\text{th}} orbit

and \lambda is the de Broglie wavelength.

For hydrogen atom, the radius of the n^{\text{th}} orbit is:

r_n = n^2 a_0

Given a_0 = 0.052 \ \mathrm{nm} and n = 2,

r_2 = 2^2 \times 0.052 = 0.208 \ \mathrm{nm}

Now substitute into the standing wave condition:

\lambda = \frac{2 \pi r_2}{n} \lambda = \frac{2 \pi \times 0.208}{2}

Simplifying,

\lambda = \pi \times 0.208 \lambda \approx 3.14 \times 0.208 \approx 0.65 \ \mathrm{nm}

This value is close to 0.67 nm.

Hence, the de Broglie wavelength of the electron is approximately 0.67 nm,

which corresponds to Option (2).