Which transition in the hydrogen spectrum wou | vedclass

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

Question

(A) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $He^

Vedclass · Accepted Answer

$n = 2$ to $n = 1$

Vedclass · Answer

(A) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $He^+$ $(Z = 2)$,the transition from $n_2 = 4$ to $n_1 = 2$ is:$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = R \left( \frac{3}{4} \right)$.For Hydrogen $(Z = 1)$,we want the transition from $n_2$ to $n_1$ such that:$\frac{1}{\lambda} = R (1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{3}{4} \right)$.Comparing the two,we need $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} = 1 - \frac{1}{4} = \frac{1}{1^2} - \frac{1}{2^2}$.Thus,the transition is from $n = 2$ to $n = 1$.

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n = 4$ to $n = 2$ of $He^+$ spectrum?

Similar Questions

If an atom has $x$ energy levels,what will be the total number of spectral lines in its emission spectrum?

Find the wavelength in the Lyman series for $n_2 = 3$ and $n_1 = 1$.

Explain: Photoelectric effect.

Select the incorrect graph for the velocity of an electron $(e^-)$ in an orbit vs. $Z$,$\frac{1}{n}$,and $n$.

In astronomical observations,signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18} \, J$ from the radiations of $600 \, nm$,calculate the number of photons received by the detector.

Vedclass Test Series

Exam Paper Generator

Online Exam Module