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College Physics (Urone)
Found in: Page 1112
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is n for a hydrogen atom if 0.850 eV of energy can ionize it?

The value of n is 4.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of ionisation energy

The energy that is enough to separate the electron in the last shell of the atom also called valance electron is the ionization energy.

Step 2: Given information and formula to be used

Consider the hydrogen atom has 4 states.

Consider the formula for the ionization energy as follows:

\[{E_n} = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\]

Here, \[{E_n}\] is the ionization energy and n is the energy level.

Step 3: Determine the value of n for a hydrogen atom if 0.850eV of energy can ionize it. 

Substitute the values and solve for the ionization state.

\[\begin{array}{l}{E_n} = 0.85\;{\rm{eV}}\\{E_n} = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\\0.85eV = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\\{n^2} = \frac{{13.6\;{\rm{eV}}}}{{0.85\;{\rm{eV}}}}\end{array}\]

Substitute the values and solve as:

\[\begin{array}{l}{n^2} = 16\\n = \sqrt {16} \\n = \pm 4\end{array}\]

Also, n can never have a negative value.

Therefore, the n value is 4.

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