In a hydrogen atom,when an electron transitio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For a given tran

Vedclass · Accepted Answer

$81,588 \, cm^{-1}$

Vedclass · Answer

(D) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For a given transition $(n_1=2, n_2=4)$,the wave number is proportional to the square of the atomic number: $\bar{
u} \propto Z^2$.For Hydrogen $(H)$,$Z_1 = 1$. For Helium ion $(He^+)$,$Z_2 = 2$.Therefore,$\frac{\bar{
u}_{He^+}}{\bar{
u}_{H}} = \left( \frac{Z_2}{Z_1} ight)^2 = \left( \frac{2}{1} ight)^2 = 4$.Thus,$\bar{
u}_{He^+} = 4 	imes \bar{
u}_{H} = 4 	imes 20,397 \, cm^{-1} = 81,588 \, cm^{-1}$.

In a hydrogen atom,when an electron transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit,the wave number is $20,397 \, cm^{-1}$. What will be the wave number when an electron in a $He^+$ ion transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit?

Similar Questions

Neglecting the reduced mass effect,which optical transition in the $He^+$ spectrum has the same wavelength as the first Lyman transition of hydrogen ($n = 2$ to $n = 1$)?

In a hydrogen-like atom,the velocity of an electron in the second orbit is $v$. What will be the velocity of the electron in its fifth orbit?

Bohr’s second postulate implies the quantisation of:

The radius of the electron's second stationary orbit in Bohr's atom is $R$. The radius of the third orbit will be:

The radius of the first (lowest) orbit of the hydrogen atom is $a_0$. The radius of the second (next higher) orbit will be (in $a_0$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module