Question

a. What wavelength photon does a hydrogen atom emit in a $200\to 199$transition?

b. What is the difference in the wavelengths emitted in a localid="1650823513478" $199\to 2$transition and a $200\to 2$transition?

Short answer

(a) In a transition, wavelength photon a hydrogen atom emit is ${\mathrm{\lambda }}_{200\to 199}=0.362\mathrm{m}$

(b) The difference in the wave lengths are$\mathrm{\Delta \lambda }=0.368\mathrm{pm}$

Step-by-step solution

Part (a) Step 1: Given information

The wavelength of hydrogen atom is$200\to 199$

Part (a) Step 2: Finding emission spectrum wavelength

We can start with an expression for the emission spectrum wavelength.

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{n}\to \mathrm{m}}=\frac{{\mathrm{\lambda }}_0}{\frac{1}{{\mathrm{m}}^2}-\frac{1}{{\mathrm{n}}^2}} \\ {\mathrm{\lambda }}_{200\to 199}=\frac{91.18\cdot {10}^{-9}}{\frac{1}{{199}^2}-\frac{1}{{200}^2}}(200\to 199\text{transition}) \\ {\mathrm{\lambda }}_{200\to 199}=0.362\mathrm{m} \end{gathered}$$

Part (b) Step 1: Given information

The given wavelengths are$199\to 2$and$200\to 2$

Part (b) Step 2: Finding difference between wavelengths

In the same approach, we can figure out $199\to 2$transition and $200\to 2$transition

localid="1651154009746" $$\begin{gathered} \begin{array}{r}{\mathrm{\lambda }}_{199\to 2}=\frac{\left(91.18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{{199}^2}}\\ {\mathrm{\lambda }}_{200\to 2}=\frac{\left(91.18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{{200}^2}}\end{array} \\ \mathrm{\Delta \lambda }={\mathrm{\lambda }}_{199\to 2}-{\mathrm{\lambda }}_{200\to 2}\text{(difference between wavelengths)} \\ \mathrm{\Delta \lambda }=\frac{\left(91.18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{{199}^2}}-\frac{\left(91.18\times {10}^{-9}\right)}{\frac{1}{2^2}-\frac{1}{{200}^2}}\text{(substitute)} \\ \mathrm{\Delta \lambda }=0.368\mathrm{pm} \end{gathered}$$