For a H atom,the wavelength for the highest e | vedclass

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

Question

(B) The Rydberg formula for wavelength is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the highest energy transition

Vedclass · Accepted Answer

$22.8$

Vedclass · Answer

(B) The Rydberg formula for wavelength is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the highest energy transition in $H$ atom $(n_1=1, n_2=\infty)$,$\lambda_H = 91.2 \ nm$.For $He^{+}$,$Z=2$. The transition corresponds to the same energy levels $(n_1=1, n_2=\infty)$.Thus,$\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{91.2}{2^2} = \frac{91.2}{4} = 22.8 \ nm$.

For a $H$ atom,the wavelength for the highest energy transition is $91.2 \ nm$. Calculate the corresponding wavelength for $He^{+}$ in $nm$.

Similar Questions

When the frequency of light incident on a metallic plate is doubled,the $KE$ of the emitted photoelectrons will be:

Wave number of a spectral line for a given transition is $y \ cm^{-1}$ for $He^{+}$,then its value for $Li^{2+}$ for the same transition is

Find the wave number of the shortest wavelength in the Balmer series.

Which of the following orders is $NOT$ correct for the $1^{st}$,$2^{nd}$,and $3^{rd}$ orbits of a hydrogen-like atom?

How did Niels Bohr obtain detailed information regarding the structure of atoms and molecules?

Vedclass Test Series

Exam Paper Generator

Online Exam Module