Question

The value of ‘n’ of the highest excited state that an electron of hydrogen atom in the ground state can reach when 12.09 eV energy is given to the hydrogen atom is.​

Answer 1

Answer:

This is your answer .....

Answer 2

Answer:

The value of 'n' is 3.

Explanation:

To find out the value of 'n' of the highest excited state that an electron of a hydrogen atom in the ground state can reach when 12.09 eV is given to the hydrogen atom, we use Rydberg's formula.

$E=E_{o}*Z^{2} (\frac{1}{n^{2} _{1} }-\frac{1}{n^{2} _{2} })$

where,

E- Energy given

E₀- Rydberg's constant (13.6eV)

Z- Atomic number

n₁- Initial level

n₂- Final level

The values we have:

E- 12.09 eV

Atomic number- 1

Initial orbit- 1

Now substitute these values:

$12.09 = 13.6 * (1)^{2} * (\frac{1}{1^{2} }-\frac{1}{n^{2} })$

$\frac{12.09}{13.6}= \frac{1}{1}-\frac{1}{n^{2} }$

$0.89 = 1 - \frac{1}{n^{2} }$

$\frac{1}{n^{2} }= 0.11$

$\frac{1}{0.11}= n^{2}$

$9 = n^{2}\\ \sqrt{9}= n\\ 3 = n$

Conclusion:

The highest level 'n' obtained is 3.