Question

For a hydrogen-like atom, the frequency of transition from $n=3$ to $n=1$ is $192\times {10}^{15}Hz$, then find the frequency of transition from $n=2$ to $n=1$?

• A. $162\times {10}^{15}Hz$
• B. $226\times {10}^{15}Hz$
• C. $226\times {10}^{18}Hz$
• D. None

Answer 1

The correct option is A

$162\times {10}^{15}Hz$

Explanation for the correct options:

A)$162\times {10}^{15}Hz$

Step 1: Frequency of transition =$=192\times {10}^{15}Hz$ ,( $n_1=1$, $n_2=3$) for ${\nu }_1$ _,( $n_1=1$, $n_2=2$) for ${\nu }_2$

Step 2: $E=hc/\lambda$

$E=h\nu$

$1/\lambda$$={\mathrm{RZ}}^2$($1/{n_1}^2$$-$ $1/{n_2}^2$)

$E=hc\left[R{Z}^2\right(1/{n_1}^2-{n_1}^2\left)\right]$

$\mathrm{h\nu }=\mathrm{hc}[{\mathrm{RZ}}^2(1/{{\mathrm{n}}_1}^2-1/{{\mathrm{n}}_2}^2\left)\right]$

$\mathrm{\nu }=\mathrm{c}[\mathrm{RZ}(1/{{\mathrm{n}}_1}^2-1/{{\mathrm{n}}_2}^2\left)\right]$

$\mathrm{\nu \alpha }=(1/{{\mathrm{n}}_1}^2-1/{{\mathrm{n}}_1}^2)$

Step 3:

Now, for transition from $\mathrm{n}=3$ to $\mathrm{n}=1$

${\mathrm{\nu }}_1$= ($1$-$1$$/$$9$ )

Now, for transition from $\mathrm{n}=2$ to $\mathrm{n}=1$

${\mathrm{\nu }}_2=$($1$-$1$$/4$ )

${\mathrm{\nu }}_2/{\mathrm{v}}_1=$ ($1$-$1$$/9$ ) $/$($1$-$1$$/4$ )

${\mathrm{\nu }}_2/{\mathrm{v}}_1=$ $3/4\times 9/8$

${\mathrm{\nu }}_2/{\mathrm{v}}_1=$ $3/4\times 9/8$$=$$27/32$

${\mathrm{\nu }}_2/{\mathrm{v}}_1=$ $27/32$$\times 192\times {10}^{15}Hz$

${\mathrm{\nu }}_2/{\mathrm{v}}_1=$ $162\times {10}^{15}Hz$

The frequency of transition from $n=2$ to $n=1$ is $162\times {10}^{15}Hz$

Explanation for the incorrect options:
Option B ,C , D are incorrect as the frequency of transition is $162\times {10}^{15}Hz$