Question 12: Atoms & Nuclei

Question

Consider a hydrogen atom with its electron in the $n$th orbit. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV , then the value of $n$ is ( $h c=1242 \mathrm{eV} \mathrm{nm}$ )

Accepted Answer

Kinetic energy of ejected electron $=$ Energy of incident photon - Energy required to ionize the electron from $n$th orbit (all in eV )
$$ \begin{array}{ll} 	herefore & 10.4=\frac{1242}{90}-\left|E_night| \ & 10.4=13.8-\frac{13.6}{n^2} \ \Rightarrow & \frac{13.6}{n^2}=3.4 \ & \left(	ext { as } E_n \propto \frac{1}{n^2} 	ext { and } E_1=-13.6 \mathrm{eV}ight) \end{array} $$
Solving this equation, we get
$$ n=2 $$

Answer

1

Answer

2

Answer

3

Answer

4

Atoms & Nuclei

Consider a hydrogen atom with its electron in the nth orbit. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV , then the value of n is ( h c=1242 \mathrm{eV} \mathrm{nm} )

Choose the correct answer: