The energy required for the ionization of an | vedclass

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

Question

(C) The ionization energy $(IE)$ of an atom is the minimum energy required to remove an electron from a given state to infinity $(n = \infty)$.
For a

Vedclass · Accepted Answer

$ < 13.6$

Vedclass · Answer

(C) The ionization energy $(IE)$ of an atom is the minimum energy required to remove an electron from a given state to infinity $(n = \infty)$.
For a hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
The ionization energy required from the $n^{th}$ state is $IE_n = E_{\infty} - E_n = 0 - (-\frac{13.6}{n^2}) = \frac{13.6}{n^2} \ eV$.
For the ground state $(n=1)$,$IE_1 = 13.6 \ eV$.
For any excited state $(n > 1)$,the value of $n^2$ is greater than $1$.
Therefore,$IE_n = \frac{13.6}{n^2} < 13.6 \ eV$ for all $n > 1$.

The energy required for the ionization of an excited hydrogen atom would be (in $eV$)-

Similar Questions

The longest wavelength doublet absorption transition is observed at $589 \, nm$ and $589.6 \, nm$. Calculate the frequency of each transition and the energy difference between the two excited states.

If the energy of a photon is $14 \, eV$ and it interacts with a hydrogen atom,which of the following is correct?

The energy of a photon is $3 \times 10^{-12} \ erg$. What is its wavelength in $nm$? $(h = 6.62 \times 10^{-27} \ erg \cdot s; \ c = 3 \times 10^{10} \ cm/s)$

For the Paschen series, the values of $n_1$ and $n_2$ in the expression $\Delta E = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ are:

Which of the following provisions of the Bohr atomic model of hydrogen turned out to be false?

Vedclass Test Series

Exam Paper Generator

Online Exam Module