Question

2. The frequency of revolution of the electron in second bohr orbit of the hydrogen atom is 8.158×10^14 hz. What is the frequency of revolution of the electron in fourth bohr orbit of the hydrogen atom?
a. 1.012×10^14 hz
b. 1.02×10^-14 hz
c. 1.012×10^10 hz
d. 1.012×10^-10 hz​

Answer 1

Given:

Frequency of revolution of electron in second Bohr orbital of hydrogen = 8.158×10^14

To find:

The frequency of revolution of the electron in fourth Bohr orbit of the hydrogen atom.

Solution:

The formula to find frequency of revolution in a particular Bohr orbital is $6.55*10^{15} *(\frac{z^{2} }{n^{3} } )$

Here z=atomic number of the element

n=orbit number

Thus the frequency of revolution of the electron in fourth Bohr orbit of the hydrogen atom will be = $6.55*10^{15} *(\frac{1^{2} }{4^{3} } )$

$=1.02*10^{14}$

Thus, option (a) is the correct one.

Answer 2

Answer:

The correct option is option a.

Explanation:

Here we have been given the frequency of revolution of the electron of the hydrogen atom in the second orbit of Bohr is

8.158 × $10^{14}$ Hz.

We have to find the frequency of the revolution in the fourth orbit of Bohr of the hydrogen atom.

Now as we know that the frequency of the particular atom in the nth orbit of Bohr is given by the following formula:

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

Where we have Z to be the atomic number of the atom, n is the orbit number and f is the frequency of the revolution of the electron in that orbit.

It is also given that at n = 2 the frequency of revolution is as follows:

$f_{2} =$ 6.55 × $10^{15}$ × $\frac{1^{2} }{2^{3} }$

= 8.158 × $10^{14}$ Hz (given)

Therefore for the fourth orbit that is n = 4 the frequency of revolution is as follows:

For hydrogen Z = 1

∴ $f_{4}$ =  6.55 × $10^{15}$ × $\frac{1^{2} }{4^{3} }$

⇒ $f_{4}$ = 6.55 × $10^{15}$ × $\frac{1^{2} }{(2^{2})^{3} }$

⇒ $f_{4}$ = 6.55 × $10^{15}$ × $(\frac{1^{2} }{2^{3} } )(\frac{1}{8} )$

⇒ $f_{4} = \frac{f_{2} }{8}$

⇒ $f_{4} = \frac{8.158 (10^{14} ) }{8}$

⇒ $f_{4}$ = 1.01975 × $10^{14}$ Hz

This is approximately equal to option a) 1.012 × $10^{14} Hz$

Hence the correct option is option a.