If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency (R = Rydberg constant, C = velocity of light)

Question

If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency (R = Rydberg constant, C = velocity of light) (A)  (3RC/27) (B)  (RC/25) (C)  (8RC/9) (D)  (5RC/36)

Tardigrade Admin · Accepted Answer

Given, n1=2 ; n2=3 Now, we know that, frequency of an emitted electron, v =R c((1/n12)-(1/n22)) =Rc((1/22)-(1/32))=R c((1/4)-(1/9)) =(5 R C/36)

Q. If an electron in hydrogen atom jumps from an orbit of level $n = 3$ to an orbit of level $n = 2$ , the emitted radiation has a frequency ( $R$ = Rydberg constant, $C$ = velocity of light)

$\frac{3 RC}{27}$

$\frac{RC}{25}$

$\frac{8 RC}{9}$

$\frac{5 RC}{36}$

Given, $n_{1} = 2; n_{2} = 3$
Now, we know that, frequency of an emitted electron,
$v = R c (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$= R c (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = R c (\frac{1}{4} - \frac{1}{9})$
$= \frac{5 RC}{36}$

Q. If an electron in hydrogen atom jumps from an orbit of level n=3 to an orbit of level n=2, the emitted radiation has a frequency (R = Rydberg constant, C = velocity of light)

273RC​

25RC​

98RC​

365RC​