A photon with energy 2.28 eV is absorbed by a

Priscilla Johnston

Priscilla Johnston

Answered question

2021-12-22

A photon with energy 2.28 eV is absorbed by a hydrogen atom. Find (a) the minimum n for a hydrogen atom that can be ionized by such a photon and (b) the speed of the electron released from the state in part (a) when it is far from the nucleus.

Answer & Explanation

boronganfh

boronganfh

Beginner2021-12-23Added 33 answers

a) Hydrogen energies are given by equation
En=13.606eVn2
Since the photon's is 2.28eV we need to find n for which
En2.28eV
We immediately see that this is not satisfied neither for n=1 nor for n=2. Let's try with n=3:
E3=13.602eV32=1.51eV
b) Since it costs 1.51eV to free this electron its kinetic energy equals
K=Ephon|E3|=2.28eV1.51eV=0.77eV
This allows us to find its speed:
v=2Kme=20.771.601019J9.111031kg
v=5.2105ms
v=520kms

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