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Found in: Page 1217

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# For what value of the principal quantum number n would the effective radius, as shown in a probability density dot plot for the hydrogen atom, be 1.00 mm? Assume that has its maximum value of n-1. (Hint:See Fig.39-24.)

The value of principal quantum number is $4.3×{10}^{3}$.

See the step by step solution

## Step 1: Identification of the given data:

The given data is listed below as,

• Radius of hydrogen atom is $r=1.00\mathrm{mm}=1.00×{10}^{-3} \mathrm{m}$

## Step 2: The principal Quantum number:

The principal quantum number is used to describe the electron’s state and is the one four quantum number assigned to each electron in an atom.

The value of the principal quantum number is natural number.

## Step 3: Determine the value of principal quantum number:

According to the fig. 39-24. the principal quantum number satisfies,

$\mathrm{r}={\mathrm{n}}^{2}\mathrm{a}$

Here, the Bohr radius is $a=5.29×{10}^{-13} m$.

Solving the above equation will give the value of the principal quantum number is,

$n=\sqrt{\frac{r}{a}}$

Substitute $1.00×{10}^{-3}\mathrm{m}$ for r in the above equation.

$\mathrm{n}=\sqrt{\frac{1.00×{10}^{-3}\mathrm{m}}{5.29×{10}^{-13}\mathrm{m}}}\phantom{\rule{0ex}{0ex}}=4.3×{10}^{3}$

Hence, the value of principal quantum number is $4.3×{10}^{3}$.