Question

Radius of first bohr orbit is 0.53 A and radius of nth bohr orbit is 212 A the value of n?

Answer 1

Answer:

2

Explanation:

ANSWER

rn=a0n2

2.12×10−19=90n2

=0.53n2

n2=4

n=2

This value of principle quantum number of excited state in 2

Answer 2

The radius of 20 th Bohr's orbit is 212 Angstrom. (n=20)

Given:
Radius of first Bohr orbit $=0.53 A$

Radius of $nth$ Bohr orbit $=212A$

To find:

The value of $n$

Solution:
Bohr's model of hydrogen atom satisfactorily provides the radius of the orbit in which the electron is revolving. This equation for Bohr's radius for Hydrogen like elements is given by

$R=\frac{n^{2} .h^{2} }{4.\pi ^{2} .m.Z.e^{2} }$

$n=$ orbit number (Principal quantum number).

$h=$ Planck's constant.

$m=$ Mass of electron.

$Z=$ Atomic number

$e=$ Charge on the electron.

From the equation, we can see that the Bohr's radius is directly proportional to the orbit in the electron is revolving that is

        $R$ ∝ $n^{2}$

Now, according t the the question, we have been given that

$R_{1}$ ∝ $(1)^{2}= 0.53 A$

$R_{n}$ ∝ $n^{2} =212A$

Taking the ratios from both sides, we see that

$\frac{n^{2} }{1}= \frac{212}{0.53}$

$n^{2} =400$

$n=20$

Final answer:

Hence, the radius of 20 th Bohr's orbit is 212 Angstrom.