Question

A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.

Answer 1

Answer:

Wavelength = 4.863 multipied by 10^-5

Explanation:

Answer 2

Explanation:

Three wavelengths can observed here.

• For the transition to (n = 3) from (n = 4) state
• For the transition to (n = 2) from (n = 3) state
• For the transition to (n = 2) from (n = 4) state

When there is a transition of the atom to (n = 2) state from (n = 4) state, let (λ1) be the wavelength

Here,

n1 = 2

n2 = 4

Now, let the wavelength be  $\frac{1}{\lambda_{1}}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$R=1.097 \times 107 m^{-1}$

$\frac{1}{\lambda_{1}}=1.097 \times 10^{7} \times\left(\frac{1}{4}-\frac{1}{16}\right)$

$\Rightarrow \frac{1}{\lambda_{1}}=\frac{1.097 \times 10^{7} \times 3}{16}$

$\Rightarrow \lambda_{1}=\frac{16 \times 10^{-7}}{3 \times 1.097}$

$=486.1 \times 10^{-9}$

$\Rightarrow$ 487 nm.

When there is a transition of the atom to (n = 3) from (n = 4), the wavelength (λ2) is shown as

Again, n1 = 3 n2 = 4

$\Rightarrow\frac{1}{\lambda_{2}}=1.097 \times 10^{7}\left(\frac{1}{9}-\frac{1}{16}\right)$ = $\frac{1}{\lambda_{2}}=1.097 \times 10^{7}\left(\frac{16-9}{144}\right)$

$\Rightarrow \lambda 2=144 / 1.097 \times 10^{7}=1875 n m$.

Likewise, for the transition to (n = 2) from (n = 3), the wavelength (λ3) is shown as

$\Rightarrow\frac{1}{\lambda_{3}}=1.097 \times 10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)$

$\Rightarrow \lambda_{3}=\frac{36 \times 10^{-7}}{1.097 \times 5}=656 \mathrm{nm}$