# Calculate energy of transition from n=3 to n=1. What region of electromagnetic spectrum is this radiation found?

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#### Explanation

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#### Explanation:

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Maggie Share
May 29, 2016

Since you did not specify, I'm going to assume you mean, hydrogen with a single electron.

${E}_{\text{ph}} = 1.89$ $\text{eV}$

$\lambda = 655.2$, this is in the optical (it's a red-pink)

#### Explanation:

The formula

E_"ph"=hcR((1/n_"lower")^2 - (1/n_"upper")^2)

$h c R = 13.6$ $\text{eV}$

${E}_{\text{ph}} = 13.6$ $\text{eV} \left({\left(\frac{1}{2}\right)}^{2} - {\left(\frac{1}{3}\right)}^{2}\right)$

${E}_{\text{ph}} = 13.6$ $\text{eV} \left(\frac{1}{4} - \frac{1}{9}\right)$

${E}_{\text{ph}} = 13.6$ $\text{eV} \left(\frac{5}{36}\right) = 1.89$ $\text{eV}$

${E}_{\text{ph}} = h \left(\frac{c}{\lambda}\right) = h c R \frac{5}{36}$

$\frac{1}{\lambda} = R \frac{5}{36}$

$\lambda = \frac{1}{R} \frac{36}{5}$

$\frac{1}{R}$ is a constant and $91$ $\text{nm}$

$\lambda = 655.2$, this is in the optical (it's a red-pink)

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