Question

Potential energy is - 27.2 eV in second orbit of
Het, then calculate double of total energy in first
excited state of hydrogen atom :-
(1) - 13.6 eV
(2) - 54.4 eV
137-6.8 eV
(4) - 27.2 eV​

Answer 1

Given:

Potential energy is -27.2 eV in second orbit of  He⁺

To Find:

Then calculate, double of total energy in first excited state of hydrogen atom

Solution:

We know that,

$\rm E_{n}=\dfrac{-13.6 \times Z^{2}}{n^{2}}$

Here,

• Z = Atomic number

• n = Orbit

Given that,

Potential energy is -27.2 eV in second orbit of  He⁺

$\rm Energy \ of \ Helium=\dfrac{-13.6 \times 4}{n^{2}}$

Hence,

$\rm P.E=-27.2 \ eV$

According to the Question,

We are asked to find the double of total energy in first excited state of hydrogen atom

Hence,

$\boxed{\rm P.E=-2 \times K.E}$

$\boxed{\rm T.E=2 \times P.E}$

Therefore,

$\rm T.E=\dfrac{-27.2}{2 \times 2} \ eV$

$\rm T.E=-6.8 \ eV$

Hence,

Option (3) is correct