Question

The velocity of an electron in a certain Bohr orbit of H-atom bears the ratio 1:275 to the velocity of light. The quantum number (n) of the orbit is :

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

Answer: (B)

2

According to Bohr, the velocity of electron of the $n^{th}$ stationary state for a hydrogen-like specie is expressed as :

$v_n=2.18\times {10}^6\times \frac{Z}{n}m/s$

where $Z=$atomic number and $n=$ orbit.

Given that $v_n:c=1:275$

where c=velocity of light = $3\times {10}^8$ m/s

$\therefore v_n=c/275$

or $v_n=\frac{3\times {10}^8}{275}=2.18\times {10}^6\times \frac{1}{n}$

$n=2.18\times {10}^6\times \frac{275}{3\times {10}^8}$

$n=1.998$ or $n=2$

therefore the velocity of an electron in second (n=2) Bohr orbit of H-atom bears the ratio 1:275 to the velocity of light.