Solveeit Logo
Login

Question

Question: The Bohr’s energy of a stationary state of hydrogen atom is given as \(E_{n} = \frac{- 2\pi^{2}me^{4...

The Bohr’s energy of a stationary state of hydrogen atom is given as En=−2π2me4n2h2E_{n} = \frac{- 2\pi^{2}me^{4}}{n^{2}h^{2}} . Putting the values of m and e for nth energy level which is not the correct value?

A

En=−21.8×10−19n2Jatom−1E_{n} = \frac{- 21.8 \times 10^{- 19}}{n^{2}}Jatom^{- 1}

B

En=−13.6n2eVatom−1E_{n} = \frac{- 13.6}{n^{2}}eVatom^{- 1}

C

En=−1312n2kJmol−1E_{n} = \frac{- 1312}{n^{2}}kJmol^{- 1}

D

En=−12.8×10−19n2ergatom−1E_{n} = \frac{- 12.8 \times 10^{- 19}}{n^{2}}ergatom^{- 1}

Answer

En=−12.8×10−19n2ergatom−1E_{n} = \frac{- 12.8 \times 10^{- 19}}{n^{2}}ergatom^{- 1}

Explanation

Solution

: En=−12.8×10−19n2ergatom−1E_{n} = \frac{- 12.8 \times 10^{- 19}}{n^{2}}ergatom^{- 1}