Question

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is 1) 5 (2) 2 B) 3 (4) 4​

Answer 1

Answer:

$66211 \times 616422 \times 115241 \times 1112512$

Answer 2

Answer:

Explanation:

Given An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 ev. If the stopping potential of the photo electron is 10 v, then the value of n is

Now the electron jumps from an exited state to ground state.

So in the ground state n = 1. Now there is a wavelength λ and the work function is given by φ = 2.75 eV

Stopping potential Vo = 10 V

The energy released E = work function + kinetic energy

So E = K.E + φ

E = e Vo + φ

   = e x 10 + 2.75 e

Energy E = 12.75 eV

Now we need to find the value of n

Now E = - 13.6 Z^2 / n^2 is in the ground state.

So E1 = - 13.6 eV

So if we put n = 4 and z = 1 we get

  E2 = - 13.6 x 1 / 16

         = - 0.85

Now E2 – E1 will be

  So – 0.85 – (- 13.6) = 12.75 eV

So both the energy will be equal when n = 4

Reference link will be

brainly.in/question/16271581

Explanation:

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