Question

What is the frequency of revolution of electron present in 2nd Bohr's orbit of H-atom.
a)1.016 x 10^16/s
b)4.065×10^16/s
c)1.626×10^15/s
d)8.2×10^14/s

[Ans:(d)]

Answer 1

the frequency of revolution of electron present in 2nd Bohr's orbit of H atom

ans is d

Answer 2

Answer : The correct option is, (d) $8.2\times 10^{14}s^{-1}$

Solution :

As we know that,

$f=\frac{v}{2\pi r}$       ................(1)

or,

$r_n=0.53\times 10^{-10}\times \frac{n^2}{Z}$      ..........(2)

$v_n=2.165\times 10^6\times \frac{Z}{n}$            ...........(3)

Now putting equation 2 and 3 in equation 1, we get

$f=(6.55\times 10^{15})\times \frac{Z^2}{n^3}$

where,

f = frequency of revolution

Z = atomic number of hydrogen atom = 1

n = number of orbit = 2

Now put all the given values in this formula, we get the frequency of revolution.

$f=(6.55\times 10^{15})\times \frac{(1)^2}{(2)^3}=8.2\times 10^{14}s^{-1}$

Therefore, the frequency of revolution is, $8.2\times 10^{14}s^{-1}$