The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?

  • A
    $4$
  • B
    $2$
  • C
    $16$
  • D
    $3$

Explore More

Similar Questions

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be .....$V$ (Take $hc = 12420\ eV\cdot\mathring{A}$).

Difficult
View Solution

The following parameter is the same for all hydrogen-like atoms and ions in their ground state.

The energy equivalent to the Rydberg constant is ...... $eV$.

In a hydrogen atom,an electron excites from the ground state to a higher energy state,and its orbital velocity is reduced to $\frac{1}{3}$ of its initial value. The radius of the orbit in the ground state is $R$. The radius of the orbit in that higher energy state is: (in $R$)

The de Broglie wavelength of an electron in the first orbit of Bohr's model is . . . . . . .

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo