Which of the following transitions has the mi | vedclass

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

Question

(A) The wavelength $(\lambda)$ of a transition is inversely proportional to the energy difference $(\Delta E)$ between the energy levels,given by the

Vedclass · Accepted Answer

$n_4 	o n_1$

Vedclass · Answer

(A) The wavelength $(\lambda)$ of a transition is inversely proportional to the energy difference $(\Delta E)$ between the energy levels,given by the relation $\Delta E = \frac{hc}{\lambda}$.For a hydrogen-like atom,the energy difference is $\Delta E = 13.6 Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.To have the minimum wavelength,the transition must have the maximum energy difference.Comparing the transitions:$(A)$ $n_4 	o n_1$: $\Delta E \propto (1 - \frac{1}{16}) = \frac{15}{16} = 0.9375$$(B)$ $n_2 	o n_1$: $\Delta E \propto (1 - \frac{1}{4}) = \frac{3}{4} = 0.75$$(C)$ $n_4 	o n_2$: $\Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$$(D)$ $n_3 	o n_1$: $\Delta E \propto (1 - \frac{1}{9}) = \frac{8}{9} \approx 0.888$Since the transition $n_4 	o n_1$ has the largest energy difference,it corresponds to the minimum wavelength.

Which of the following transitions has the minimum wavelength?

Similar Questions

Assertion : For Balmer series of hydrogen spectrum,the value $n_1 = 2$ and $n_2 = 3, 4, 5...$
Reason : The value of $n$ for a line in Balmer series of hydrogen spectrum having the highest wavelength is $4$ and $6$.

Which of the following species has the same radius as that of the hydrogen atom in the $n = 1$ state?

What is the difference between a photon and a quantum?

The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6 \ eV$. The energy value of an electron in the first excited state of $Li^{2+}$ is ............... $eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module