Question

If the energy in the first excited state in hydrogen atom is $23.8eV$ then the potential energy of a hydrogen atom in the ground state can be assumed to be

• A. $10eV$
• B. $23.3eV$
• C. $-13.6eV$
• D. Zero

Answer 1

Answer: (B)

$23.3eV$

Given that,

Energy in the first excited state $=23.8eV$

Now, we know that,

Total energy of any level in hydrogen atom is $E=-\frac{13.6}{n^2}eV$

For ground state

$E_g=-13.6eV$

Now, for exited state

$E_e=-\frac{13.6}{4}eV$

$E_e=-3.4eV$

Now, the potential of ground state

$P.E=2\times (-13.6)$

$P.E=-27.2eV$

Now, if we assume, it is zero now means we increase hydrogen energy level by 27.2 eV energy

So, now in first excited state it has total energy of

$=-3.4+27.2$

$=23.8eV$

Hence, this is the required solution