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

Q.27) De-Broglie wavelength of an electron orbiting in the $n=2$ state of hydrogen atom is close to $($ Given Bohr radius $=0.052 \mathrm{~nm})$

A) 2.67 nm

B) 0.167 nm

C) 0.67 nm

D) 1.67 nm

Answers (1)

best_answer

Solution:

Correct Option: (3) 0.67 nm

In the Bohr model, the radius of the $n$-th orbit is given by $r_n=n^2 a_0$. The de-Broglie wavelength is then:

$$
\lambda=\frac{2 \pi r_n}{n}=\frac{2 \pi n^2 a_0}{n}=2 \pi a_0 n
$$


Substituting $a_0=0.052 \mathrm{~nm}$ and $n=2$ :

$$
\lambda=2 \cdot \pi \cdot 0.052 \cdot 2=0.104 \cdot \pi \approx 0.104 \cdot 3.14 \approx 0.65 \mathrm{~nm}
$$


Which is approximately 0.67 nm .
Hence, the answer is option (3).

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