Electron Energy and Electron Energy Levels in Hydrogen Atom Questions in English

(C) The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{-13.6 \text{ eV}}{n^2}$.
For the first orbit $(n=1)$,the energy is $E_1 = -13.6 \text{ eV}$.
For the second orbit $(n=2)$,the energy is $E_2 = \frac{-13.6 \text{ eV}}{2^2} = -3.4 \text{ eV}$.
The minimum excitation energy is the energy required to move the electron from the ground state $(n=1)$ to the first excited state $(n=2)$.
$\Delta E = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}$.

(B) For a hydrogen atom,the energy of the ground state $(n=1)$ is $E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV}$.
For the second excited state,$n=3$. The energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}$.
The energy required to excite the electron from the ground state to the second excited state is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \approx 12 \text{ eV}$.
$(b)$ The ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$.
$E_{\infty} = 0 \text{ eV}$.
Energy required = $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \text{ eV}$.

(C) The total energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} eV$.
For the excited state $n=4$,the total energy is $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 eV$.
The potential energy $(PE)$ of an electron in a Bohr orbit is related to the total energy $(E)$ by the relation $PE = 2E$.
Therefore,$PE = 2 \times (-0.85 eV) = -1.7 eV$.

(C) The energy of the $n^{th}$ level in a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The energy difference between consecutive levels is $\Delta E = E_{n+1} - E_n = 13.6 \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right]$.
Simplifying this expression,we get $\Delta E = 13.6 \left[ \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right] = 13.6 \left[ \frac{2n+1}{n^2(n+1)^2} \right]$.
For large values of $n$,$\Delta E \approx 13.6 \left[ \frac{2n}{n^4} \right] = \frac{27.2}{n^3}$.
Since $\Delta E \propto \frac{1}{n^3}$,as the quantum number $n$ increases,the energy difference $\Delta E$ decreases.

(D) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final}$.
Since the energy levels are $E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$,the energy difference for a transition from $n_i$ to $n_f$ is $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Emission occurs when an electron moves from a higher energy level to a lower energy level (downward arrows).
Looking at the diagram:
Transition $(II)$ is from $n=4$ to $n=3$.
Transition $(III)$ is from $n=2$ to $n=1$.
Transition $(IV)$ is from $n=3$ to $n=2$.
Transition $(I)$ is an absorption (upward arrow),so it does not represent emission.
Comparing the energy differences:
For $(II)$: $\Delta E \propto (\frac{1}{3^2} - \frac{1}{4^2}) = (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.0486$.
For $(III)$: $\Delta E \propto (\frac{1}{1^2} - \frac{1}{2^2}) = (1 - 0.25) = 0.75$.
For $(IV)$: $\Delta E \propto (\frac{1}{2^2} - \frac{1}{3^2}) = (\frac{1}{4} - \frac{1}{9}) = \frac{5}{36} \approx 0.1389$.
Comparing the values,the transition $(III)$ has the largest energy difference,corresponding to the emission of the most energetic photon.

(B) The minimum excitation energy corresponds to the transition from the ground state $(n=1)$ to the first excited state $(n=2)$.
Using the formula for energy levels of the Hydrogen atom: $E_n = -\frac{13.6}{n^2} \ eV$.
The energy of the ground state is $E_1 = -13.6 \ eV$.
The energy of the first excited state is $E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.
The excitation energy required is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
Therefore,the minimum excitation potential is $10.2 \ V$.

(B) The energy of the emitted photon is given by $\Delta E = 13.6 \ eV \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $n_2=4$ to $n_1=2$,$\Delta E = 13.6 \times \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \times \frac{3}{16} = 2.55 \ eV$.
Converting to Joules: $\Delta E = 2.55 \times 1.6 \times 10^{-19} \ J = 4.08 \times 10^{-19} \ J$.
The momentum of the photon is $p = \frac{E}{c} = \frac{4.08 \times 10^{-19}}{3 \times 10^8} = 1.36 \times 10^{-27} \ kg \ m \ s^{-1}$.
By conservation of momentum,the recoil momentum of the hydrogen atom is equal to the momentum of the photon.
$m_{H} v = p$,where $m_{H} \approx 1.67 \times 10^{-27} \ kg$.
$v = \frac{1.36 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 0.814 \ m \ s^{-1}$.

(D) Let the ground state energy of the hydrogen-like atom be $-E$.
For the first excitation,the electron transitions from the ground state $(n=1)$ to the first excited state $(n=2)$.
The energy required is $E_{2} - E_{1} = 15 \ eV$.
$\left(\frac{-E}{2^2}\right) - \left(\frac{-E}{1^2}\right) = 15 \ eV$.
$\frac{-E}{4} + E = 15 \ eV$.
$\frac{3E}{4} = 15 \ eV \implies E = 20 \ eV$.
For the third excitation,the electron transitions from the ground state $(n=1)$ to the fourth excited state $(n=4)$.
The energy required is $E_{4} - E_{1} = \left(\frac{-E}{4^2}\right) - \left(\frac{-E}{1^2}\right)$.
$= \frac{-E}{16} + E = \frac{15E}{16}$.
Substituting $E = 20 \ eV$,we get:
$E_{3rd\ excitation} = \frac{15 \times 20}{16} = \frac{300}{16} = \frac{75}{4} \ V$.
Thus,the correct option is $D$.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Here,$E = 5 \ eV$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $1 \ eV = 1.6 \times 10^{-19} \ J$.
Alternatively,using the shortcut formula $E \ (eV) = \frac{1240}{\lambda \ (nm)}$.
Substituting the values: $5 = \frac{1240}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{1240}{5} = 248 \ nm$.
Therefore,the wavelength of the emitted photon is $248 \ nm$.

(A) The energy of a photon emitted during a transition is given by $E = \Delta E = 3.31 \ eV$.
The relationship between energy and wavelength is $E = \frac{hc}{\lambda}$.
Given $h = 6.62 \times 10^{-34} \ J \cdot s$, $c = 3 \times 10^8 \ m/s$, and $1 \ eV = 1.6 \times 10^{-19} \ J$.
Using the formula $\lambda = \frac{hc}{E}$, we have:
$\lambda = \frac{(6.62 \times 10^{-34} \ J \cdot s) \times (3 \times 10^8 \ m/s)}{3.31 \times 1.6 \times 10^{-19} \ J}$.
$\lambda = \frac{19.86 \times 10^{-26}}{5.296 \times 10^{-19}} \ m$.
$\lambda \approx 3.75 \times 10^{-7} \ m$.
Converting to $\text{\AA}$ $(1 \ Å = 10^{-10} \ m)$:
$\lambda \approx 3750 \ Å$.

(A) The energy of an electron in the $n$-th state of a hydrogen atom is given by $E_n = -\frac{|E|}{n^2}$.
Given,the initial state is $n_A = 3$,so the initial energy is $E_{n_A} = -\frac{|E|}{3^2} = -\frac{|E|}{9}$.
When a photon of energy $\Delta E = \frac{|E|}{12}$ is absorbed,the electron transitions to state $n_B$ with energy $E_{n_B} = -\frac{|E|}{n_B^2}$.
The energy conservation equation is $E_{n_B} - E_{n_A} = \Delta E$.
Substituting the values: $-\frac{|E|}{n_B^2} - (-\frac{|E|}{9}) = \frac{|E|}{12}$.
Dividing by $|E|$: $-\frac{1}{n_B^2} + \frac{1}{9} = \frac{1}{12}$.
Rearranging the terms: $\frac{1}{n_B^2} = \frac{1}{9} - \frac{1}{12}$.
Finding a common denominator: $\frac{1}{n_B^2} = \frac{4 - 3}{36} = \frac{1}{36}$.
Therefore,$n_B^2 = 36$,which gives $n_B = 6$.

(C) The kinetic energy $(K)$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $K_n = \frac{13.6 \text{ eV}}{n^2}$.
For the third excited state,the principal quantum number is $n_1 = 3 + 1 = 4$.
For the fourth excited state,the principal quantum number is $n_2 = 4 + 1 = 5$.
The kinetic energy in the third excited state is $K_4 = \frac{13.6}{4^2} = \frac{13.6}{16}$.
The kinetic energy in the fourth excited state is $K_5 = \frac{13.6}{5^2} = \frac{13.6}{25}$.
The ratio of the kinetic energies is $\frac{K_4}{K_5} = \frac{13.6/16}{13.6/25} = \frac{25}{16}$.
Thus,the ratio is $25: 16$.

(C) The energy required to ionize a hydrogen atom from its ground state $(n=1)$ is equal to the ionization energy, which is $E = 13.6 \, eV$.
To find the maximum wavelength $(\lambda_{max})$ of the incident radiation, we use the relation $E = \frac{hc}{\lambda}$.
Substituting the values: $13.6 \, eV = \frac{1240 \, eV \cdot nm}{\lambda}$.
$\lambda = \frac{1240}{13.6} \, nm \approx 91.17 \, nm$.
Converting this to $\mathring{A}$ $(Å)$: $91.17 \, nm = 911.7 \, Å \approx 912 \, Å$.
Thus, the maximum wavelength required is nearly $912 \, Å$.

(A) For a hydrogen atom,the total energy in the $n^{th}$ orbit is given by $E_n = \frac{E_1}{n^2}$,where $E_1 = -13.6 \ eV$.
For the first excited state,$n = 2$.
Therefore,the total energy $E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \ eV$.
The relationship between total energy $(E)$ and potential energy $(U)$ is $U = 2E$.
Thus,the potential energy in the first excited state is $U_2 = 2 \times E_2 = 2 \times (-3.4 \ eV) = -6.8 \ eV$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
For the ground state,$n = 1$.
The first excited state corresponds to $n = 2$.
The second excited state corresponds to $n = 3$.
We need the ratio of energies in the first excited state $(E_2)$ and the second excited state $(E_3)$.
$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} \text{ eV}$.
$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV}$.
The ratio of the energy in the first excited state to the energy in the second excited state is $\frac{E_2}{E_3} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.
Thus,the ratio is $9: 4$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = \frac{-13.6}{n^2} eV$.
For the ground state,$n = 1$.
The first excited state is $n = 2$,the second excited state is $n = 3$,the third excited state is $n = 4$,and the fourth excited state is $n = 5$.
Substituting $n = 5$ into the formula:
$E_5 = \frac{-13.6}{5^2} eV = \frac{-13.6}{25} eV = -0.544 eV$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the ground state $(n_1 = 1)$, the energy is $E_1 = -13.6 \text{ eV}$.
For the excited state $(n_2 = 3)$, the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}$.
The energy required to excite the atom from $n=1$ to $n=3$ is $\Delta E = E_3 - E_1$.
$\Delta E = -1.51 - (-13.6) = 12.09 \text{ eV}$.
Rounding this value, we get $\Delta E \approx 12.1 \text{ eV}$.

(D) The energy of the excited state is given by the sum of the ground state energy and the excitation energy.
$E_{\text{excited}} = E_{\text{ground}} + E_{\text{excitation}} = 10.5 \text{ eV} + 7.7 \text{ eV} = 18.2 \text{ eV}$.
This represents the higher energy level $(E_1)$ involved in the transition.
The energy of the emitted photon corresponding to the wavelength $\lambda = 4960 \mathring A = 496 \text{ nm}$ is calculated as:
$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1240 \text{ eV nm}}{496 \text{ nm}} = 2.5 \text{ eV}$.
Since the photon is emitted during a transition from a higher level $(E_1)$ to a lower level $(E_2)$,we have:
$E_{\text{photon}} = E_1 - E_2$
$2.5 \text{ eV} = 18.2 \text{ eV} - E_2$
$E_2 = 18.2 \text{ eV} - 2.5 \text{ eV} = 15.7 \text{ eV}$.
Thus,the energies of the two levels are $18.2 \text{ eV}$ and $15.7 \text{ eV}$.

(D) The energy of a photon emitted during a transition in a hydrogen atom is given by the Rydberg formula: $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
For the transition $n=2 \rightarrow n=1$,the energy $E_1$ is:
$E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} \text{ eV}$.
For the transition $n=3 \rightarrow n=2$,the energy $E_2$ is:
$E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36} \text{ eV}$.
Now,calculating the ratio $E_2 / E_1$:
$\frac{E_2}{E_1} = \frac{13.6 \times (5/36)}{13.6 \times (3/4)} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{9 \times 3} = \frac{5}{27}$.

(C) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
For the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \text{ eV}$.
For the first excited state $(n_2 = 2)$,the energy is $E_2 = -\frac{13.6 \text{ eV}}{2^2} = -\frac{13.6}{4} \text{ eV} = -3.4 \text{ eV}$.
The energy of the photon released during the transition from $n_2$ to $n_1$ is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 13.6 \text{ eV} - 3.4 \text{ eV} = 10.2 \text{ eV}$.

(B) The potential energy $U$ of an electron in a hydrogen atom at a state $n$ is given by $U_n = -27.2 / n^2 \ eV$.
For the first excited state $(n = 2)$,the potential energy is $U_2 = -27.2 / 2^2 = -6.8 \ eV$.
According to the problem,we assume $U_2 = 0$. This implies we are adding a constant $C = 6.8 \ eV$ to the potential energy scale.
The total energy $E_n$ of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.
For $n = \infty$,the standard total energy is $E_{\infty} = 0 \ eV$.
Since we shifted the potential energy scale by adding $6.8 \ eV$,the new total energy $E'_{\infty}$ will be $E_{\infty} + 6.8 \ eV = 0 + 6.8 \ eV = 6.8 \ eV$.

(B) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \text{ eV}$.
The energy required to excite the atom to the $n$-th energy level is given by $\Delta E = E_n - E_1$,where $E_n = -13.6/n^2 \text{ eV}$.
For $n=2$,$\Delta E = -3.4 - (-13.6) = 10.2 \text{ eV}$.
For $n=3$,$\Delta E = -1.51 - (-13.6) = 12.09 \text{ eV}$.
For $n=4$,$\Delta E = -0.85 - (-13.6) = 12.75 \text{ eV}$.
Since the incident electron beam has an energy of $12.5 \text{ eV}$,it can provide enough energy to excite the hydrogen atoms to the $n=3$ level $(12.09 \text{ eV})$,but it does not have enough energy to reach the $n=4$ level $(12.75 \text{ eV})$.
Therefore,the hydrogen atoms will be excited up to the $n=3$ energy level.

(D) The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the $n=2$ state,the total energy is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
The relationship between potential energy $(PE)$ and total energy $(E)$ is $PE = 2E$.
Therefore,the potential energy in the $n=2$ state is $PE = 2 \times (-3.4 \ eV) = -6.8 \ eV$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
For the $M$-shell $(n=3)$,the energy is $E_3 = -\frac{13.6 \ eV}{3^2} = -\frac{13.6}{9} \ eV \approx -1.51 \ eV$.
To ionize the hydrogen atom from the $M$-state,the atom must be provided with enough energy to reach the ionization limit $(E_{\infty} = 0 \ eV)$.
The energy required is $\Delta E = E_{\infty} - E_3 = 0 - (-1.51 \ eV) = +1.51 \ eV$.
Therefore,the second electron must transfer at least $+1.51 \ eV$ of energy to the atom.