Question

A hydrogen atom in its first excited state absorbs a photon of energy x × 10^-2 eV and exited to a higher energy state where the potential energy of electron is -1.08 eV. The value of x is ______.

Options

• 286

• 386

• 186

• 486

Answer 1

A hydrogen atom in its first excited state absorbs a photon of energy x × 10^-2 eV and exited to a higher energy state where the potential energy of electron is -1.08 eV. The value of x is 286.

Explanation:

As, En = `("P"."E"_"n")/2 = -1.08/2 = -0.544`

So, ΔE, E_f - E_i = -0.544 - `(-13.6/2^2) = 3.4 - 0.544`

≈ 2.86eV = 286 × 10^-2 eV