Physics

Atoms

The Line Spectra of the Hydrogen Atom

The ground state energy of hydrogen atom is -13.6 {~eV}. The p

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Solving time: 5 mins

The ground state energy of hydrogen atom is $- 13.6 e V$ . The photon emitted during the transition of electron from $n = 3$ to $n = 1$ state, is incident on a photosensitive material of unknown work function.
The photoelectrons are emitted from the materials with a maximum kinetic energy of 9eV. Calculate the threshold wavelength of the material used.

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Text solutionVerified

Given, ground state energy

E_{1} = - 13.6 e V

Energy of electron in

n

th orbit

E_{n} = \frac{13.6}{n ^{2}} e V

For

n = 1

E_{1} = - 13.6 e V

For

n = 3

E_{3} = \frac{- 13.6}{3 ^{2}} = - 1.5 e V

∴

The energy of photon released during the transition of electron from

n = 3

n = 1

E = E_{3} - E_{1}

= - (1.5) - (- 13.6)

= 12.1 e V

Now, Einstein's photoelectric equation energy of photon

(E) = K E_{m a x} + ϕ

where,

ϕ

is work function of metal

12.1 e V = 9 e V + ϕ

ϕ = 12.1 - 9

= 3.1 e V

∴

Work function

ϕ = 3.1 e V

∴

The wavelength correspond to

3.1 e V

, threshold wavelength

λ_{0} = \frac{1242 e V - nm}{ϕ ( in e V )}

= \frac{1242 e V - nm}{3.1 e V} = 401 nm

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The ground state energy of hydrogen atom is

- 13.6 e V

. The photon emitted during the transition of electron from

n = 3

n = 1

state, is incident on a photosensitive material of unknown work function.
The photoelectrons are emitted from the materials with a maximum kinetic energy of 9eV. Calculate the threshold wavelength of the material used.

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Question Text	The ground state energy of hydrogen atom is $- 13.6 e V$ . The photon emitted during the transition of electron from $n = 3$ to $n = 1$ state, is incident on a photosensitive material of unknown work function. The photoelectrons are emitted from the materials with a maximum kinetic energy of 9eV. Calculate the threshold wavelength of the material used.
Topic	Atoms
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	28

Text solutionVerified

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