The ionization energy of hydrogen atom is $- 13.6 eV .$ The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro's constant $= 6.022 \times 10^{23})$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$1.69 \times 10^{25} J$

$1.69 \times 10^{- 20} J$

$1.69 \times 10^{- 23} J$

$1.69 \times 10^{23} J$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$E = \frac{- 13.6}{n^{2}} = \frac{- 13.6}{4} = - 3.4 eV$

We know that energy required for excitation

$Δ E = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV .$

Therefore energy required for excitation of electron per

atom $= \frac{10.2}{6.02 \times 10^{23}} = 1.69 \times 10^{- 23} J .$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE