Which of the following transitions will emit | vedclass

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

Question

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = h 
u = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Del

Vedclass · Accepted Answer

$n_4 	o n_1$

Vedclass · Answer

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = h 
u = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the shortest wavelength corresponds to the transition with the largest energy difference $(\Delta E)$.Comparing the transitions:$(A)$ $n_4 	o n_1$: Largest energy gap.$(B)$ $n_2 	o n_1$: Smaller energy gap.$(C)$ $n_4 	o n_2$: Smaller energy gap.$(D)$ $n_3 	o n_1$: Smaller energy gap than $n_4 	o n_1$.Therefore,the transition $n_4 	o n_1$ involves the maximum energy change,resulting in the shortest wavelength.

Which of the following transitions will emit the shortest wavelength?

Similar Questions

The wavelength of the first line of the Lyman series for hydrogen is $1216 \ \mathring{A}$. The wavelength of the first line of the same series for a $10$ times ionized sodium atom $(Z = 11)$ will be ......... $\mathring{A}$.

If in a hydrogen atom,an electron jumps from the $3^{rd}$ Lyman line to the $1^{st}$ Lyman line,then the obtained wavelength is:

The difference in radii between the fourth and third Bohr orbits of $He^{+}$ (in $m$) is:

The energy required to excite an electron in a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $..........$ $eV$.

What will be the number of photons required to obtain $1 \ J$ of energy by radiation having a wavelength of $4000 \ \mathring{A}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module