Structure of Atom

(i) Number of electrons in hydride ion $\left(\mathrm{H}^{-}\right)$ is $=2$

(ii) Electronic configuration of ground state in $\mathrm{H}^{-}$ ion (G.S) = 1s²

(iii) Electronic configuration of first excited state of $\mathrm{H}^{-}$ion $\left(\mathrm{E}{S_1}\right)$

(iv) Electronic configuration of second excited state of $\mathrm{H}^{-}\left(\right.$E.S $\left._2\right)$

(v) The electron in $2 p$ orbital can occupy any three $2 p$ orbitals $2 p_{x^{\prime}} 2 p_y$ and $2 p_z$ as follows :

Hence, three degenerate orbitals represents second excited state of H^–.

As n = 4, so l = 0, 1, 2, 3 which implies the orbitals are 4s, 4p, 4d and 4f.

Now, |m_l | = 1 implies m_l = +1 and −1. Therefore, l can be 3, 2, 1, as for l = 0, m_l = 0.

For l = 1, m_l = −1, 0, +1

For l = 2, m_l = −2, −1, 0, +1, +2

For l = 3, m_l = −3, −2, −1, 0, +1, +2, +3

To solve this problem, we will use the de Broglie wavelength formula, which is given by

$\lambda = \frac{h}{mv}$

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is the velocity of the particle.

However, it's more relevant to use the form of the de Broglie equation that involves temperature, given that kinetic energy ($KE$) at a particular temperature is linked to the velocity of the gas particles. The kinetic energy for a gas particle can be calculated using the equation:

$KE = \frac{3}{2}k_{B}T$

where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin. The velocity $v$ of a gas particle can be related to its kinetic energy by the equation:

$KE = \frac{1}{2}mv^2$

Solving for $v$ gives:

$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{3k_BT}{m}}$

Substituting $v$ in the de Broglie wavelength formula gives:

$\lambda = \frac{h}{m\sqrt{\frac{3k_BT}{m}}} = \frac{h}{\sqrt{3mk_BT}}$

Given that we are comparing helium (He) and neon (Ne) gases, the relative de Broglie wavelengths can be related by the mass of the particles and their temperatures. For helium and neon, respectively, this becomes:

$\lambda_{He} = \frac{h}{\sqrt{3m_{He}k_BT_{He}}}$

$\lambda_{Ne} = \frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}$

Given the temperatures are –73°C for He and 727°C for Ne, we first convert these temperatures to Kelvin:

$T_{He} = 200K \quad (\text{-73°C + 273})$

$T_{Ne} = 1000K \quad (\text{727°C + 273})$

The ratio of the de Broglie wavelength of He to Ne, $\frac{\lambda_{He}}{\lambda_{Ne}}$, would thus be:

$\frac{\lambda_{He}}{\lambda_{Ne}} = \frac{\frac{h}{\sqrt{3m_{He}k_BT_{He}}}}{\frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}} = \sqrt{\frac{m_{Ne}T_{Ne}}{m_{He}T_{He}}}$

Substituting the masses ($m_{He}=4$ a.m.u. and $m_{Ne}=20$ a.m.u., with $1$ a.m.u. approximately equal to $1.660539 \times 10^{-27}$ kg, though this conversion is not strictly necessary for a ratio where units cancel out) and the temperatures:

$\frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$

Therefore, the value of $M$, which is the multiplier for the de Broglie wavelength of He gas at –73°C compared to that of Ne at 727°C, is 5.

To answer this question, we need to consider the quantum numbers that describe electrons in atoms. Each electron in an atom is described by four quantum numbers:

1. The principal quantum number ($n$) dictates the energy level and size of the electron orbital.
2. The azimuthal (angular momentum) quantum number ($l$) defines the shape of the orbital. For a given $n$, $l$ can take any integer value from 0 to $n-1$.
3. The magnetic quantum number ($m_l$) describes the orientation of the orbital in space. For a given $l$, $m_l$ can take values from $-l$ to $+l$, including zero.
4. The spin quantum number ($m_s$) describes the direction of the electron's spin and can have a value of $+\frac{1}{2}$ or $-\frac{1}{2}$.

Given the principal quantum number, $n = 3$, and the spin quantum number, $m_s = -\frac{1}{2}$, we need to calculate the maximum number of electrons that can fit this criteria.

For $n = 3$, the possible values of $l$ are 0, 1, and 2 (s, p, and d orbitals respectively). Here's how the orbitals break down:

• When $l = 0$ (the 3s orbital), $m_l = 0$, so there's 1 orbital.
• When $l = 1$ (the 3p orbitals), $m_l$ can be -1, 0, or +1, providing 3 orbitals.
• When $l = 2$ (the 3d orbitals), $m_l$ can be -2, -1, 0, +1, or +2, giving 5 orbitals.

Each orbital can hold 2 electrons with opposite spins. Since we are specifically looking for electrons with $m_s = -\frac{1}{2}$, we can only count one electron per orbital. Therefore, we sum the total number of orbitals across the s, p, and d sublevels for $n = 3$:

3s: 1 orbital $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 1 electron

3p: 3 orbitals $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 3 electrons

3d: 5 orbitals $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 5 electrons

If we sum these, we get:

1 + 3 + 5 = 9 electrons

Thus, the maximum number of electrons that can have the principal quantum number $n = 3$ and the spin quantum number $m_s = -\frac{1}{2}$ is 9.

To determine the number of metals that will show the photoelectric effect when light of 300 nm wavelength falls on them, we first need to calculate the energy (in eV) of the incident photons. The energy ($E$) of a photon can be calculated using the formula:

$E = \frac{hc}{\lambda}$

where

• $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s),


• $c$ is the speed of light ($3.00 \times 10^{8}$ m/s),


• $\lambda$ is the wavelength of the light (in meters).

First, convert the wavelength from nm to meters:

$300\, \text{nm} = 300 \times 10^{-9}\, \text{m}$

Then, calculate the energy of the photons:

$E = \frac{(6.626 \times 10^{-34}\, \text{J·s}) \times (3.00 \times 10^{8}\, \text{m/s})}{300 \times 10^{-9}\, \text{m}}$

$E = \frac{(6.626 \times 3.00) \times 10^{-19}}{300}\, \text{J}$

$E = \frac{19.878 \times 10^{-19}}{300}\, \text{J}$

$E \approx 6.626 \times 10^{-19}\, \text{J}$

To convert the energy from joules to electron volts (eV), we divide by the charge of an electron ($1.602 \times 10^{-19}$ C):

$E \approx \frac{6.626 \times 10^{-19}\, \text{J}}{1.602 \times 10^{-19}\, \text{C/e^-}} \approx 4.14\, \text{eV}$

Now, to determine whether the photoelectric effect will occur, we compare the energy of the incident photons to the work function ($\phi$) of each metal. The photoelectric effect occurs if the photon energy is greater than or equal to the work function of the metal:

• For Li ($\phi = 2.4\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.4\, \text{eV}$


• For Na ($\phi = 2.3\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.3\, \text{eV}$


• For K ($\phi = 2.2\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.2\, \text{eV}$


• For Mg ($\phi = 3.7\, \text{eV}$): Yes, $4.14\, \text{eV} > 3.7\, \text{eV}$


• For Cu ($\phi = 4.8\, \text{eV}$): No, $4.14\, \text{eV} < 4.8\, \text{eV}$


• For Ag ($\phi = 4.3\, \text{eV}$): No, $4.14\, \text{eV} < 4.3\, \text{eV}$


• For Fe ($\phi = 4.7\, \text{eV}$): No, $4.14\, \text{eV} < 4.7\, \text{eV}$


• For Pt ($\phi = 6.3\, \text{eV}$): No, $4.14\, \text{eV} < 6.3\, \text{eV}$


• For W ($\phi = 4.75\, \text{eV}$): No, $4.14\, \text{eV} < 4.75\, \text{eV}$

Thus, the number of metals which will show the photoelectric effect when light of 300 nm wavelength falls on the metal is 4 (Li, Na, K, Mg).

According to Bohr's theory

$mvr = {{nh} \over {2\pi }}$

In the third orbit, n = 3

$\therefore$ $mvr = 3\left( {{h \over {2\pi }}} \right)$ ...... (i)

According to de Broglie expression

$\lambda = {h \over {mv}}$ ...... (ii)

Substituting this in equation (i), we get

$\left( {{h \over \lambda }} \right)r = 3\left( {{h \over {2\pi }}} \right)$ or $3\lambda = 2\pi r$

Thus, the circumference of third orbit is equal to three times the wavelength of electron i.e., the number of waves made by Bohr electron in one complete revolution is three.

In general, the number of waves made by a Bohr electron in an orbit is equal to its principal quantum number.

For n = 3 and l = 2 (i.e., 3d orbital), the values of m varies from –2 to +2, that is, −2, −1, 0, +1, +2 and for each m there are 2 values of s, that is, +1/2 and −1/2.

So the maximum number of electrons in five 3d orbitals is 5 × 2 = 10.

For hydrogen atom;

z = 1, n = 1

or, v = 2.18 $\times$ 10⁶ ms^$-$1

de-Broglie's wavelength, $\lambda = {h \over {mv}}$

$ = {{6.626 \times {{10}^{ - 34}}kg\,{m^2}{s^{ - 1}}} \over {9.11 \times {{10}^{ - 31}}kg \times 2.18 \times {{10}^6}\,m{s^{ - 1}}}}$

$ = 3.34 \times {10^{ - 10}}\,m$

Orbital angular momentum

$ = \sqrt {l(l + 1)} .\,{h \over {2\pi }}$

For 2p-orbital, l = 1

$\therefore$ Orbital angular momentum $ = \sqrt 2 \,.\,h/2\pi $.

The Bohr's model was proposed by Niels Bohr and Ernst Rutherford which states that the electrons revolve around the nucleus in circular orbits known as energy shells.

The angular momentum for an electron can be calculated using following formula,

$mvr = {{nh} \over {2\pi }}$ ..... (i)

Here, m is the mass of an electron and $v$ is the velocity of electron. The $r$ represents the radius of the $n^{th}$ orbit. The $h$ denotes Planck’s constant and $n$ denotes the number of orbits in which electron is revolving.

$r = {a_0} = 0.529\,\mathop A\limits^o = 0.529 \times {10^{ - 10}}$ m

$h = 6.626 \times {10^{ - 34}}$ kg m$^2$/s

$m = 9.1 \times {10^{ - 31}}$ kg

$n = 1$

$\pi=3.14$

Putting respective values in equation (i),

$9.1 \times {10^{ - 31}} \times v \times 0.529 \times {10^{ - 10}} = {{1 \times 6.626 \times {{10}^{ - 34}}} \over {2 \times 3.14}}$

$v = {{1 \times 6.626 \times {{10}^{ - 34}}} \over {2 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 0.529 \times {{10}^{ - 10}}}}$ m/s

$v = 2.19 \times {10^6}$ m/s

Hence, the velocity of electron in the first Bohr orbit of hydrogen atom is $2.19 \times {10^6}$ m/s.

(B) The angular momentum of an electron orbit is given as,

$mvr = {{nh} \over {2\pi }}$ ..... (i)

The de Broglie wavelength is given by,

$\lambda = {h \over p}$ ..... (ii)

Here, $\lambda$ denotes de Broglie wavelength, h is Planck’s constant and P is the momentum.

The momentum P is given as,

$P = mv$

Substitute value of P in equation (i),

$mvr = {{nh} \over {2\pi }} \Rightarrow \Pr = {{nh} \over {2\pi }}$

From equation (ii),

${h \over \lambda } = {{nh} \over {2\pi r}}$

On simplifying further,

$\lambda = {{2\pi r} \over n}$ ..... (iii)

$\lambda = 2\pi \times 0.529\,\mathop A\limits^o $

We know, for first Bohr's orbit,

n = 1 and r = 0.529 $\times$ 10$^{-10}$ m

$\lambda=2\times3.14\times0.529\times10^{-10}$ m

$=3.32\times10^{-10}$ m

1 angstrom = 10$^{-10}$ m

$\therefore$ $\lambda = 3.32\,\mathop A\limits^o $

Therefore, the de Broglie's wavelength of the electron in first Bohr orbit is 3.32 angstrom.

(C) The orbital angular momentum of an orbital in terms of $\frac{h}{2\pi}$ is given by,

Orbital angular momentum = $\sqrt {l(l + 1)} {h \over {2\pi }}$

Here, $l$ is the angular momentum quantum number.

For $2p$ orbital, $l = 1$

Substituting value of $l$ in above equation, we get

Orbital angular momentum $ = \sqrt {l(l + 1)} {h \over {2\pi }}$

Orbital angular momentum $ = \sqrt 2 {h \over {2\pi }}$

Therefore, the orbital angular momentum for 2p orbital is $\sqrt 2 {h \over {2\pi }}$.

Final Answer

(A) The velocity of electron in the first Bohr orbit of hydrogen atom is $2.19 \times {10^6}$ m/s.

(B) $\lambda = 3.32\,\mathop A\limits^o $

(C) The orbital angular momentum for 2p orbital is $\sqrt 2 {h \over {2\pi }}$.

Shortcut Method:

The de Broglie's wavelength for first Bohr orbit can be calculated as follows:

$\lambda = {h \over {mv}} = {{6.63 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times 2.18 \times {{10}^6}}}$

$ = 0.33 \times {10^{ - 9}}$ m $ = 3.3\,\mathop A\limits^o $

To find the wavelength of a ball moving with a certain velocity, we use the de Broglie wavelength formula:

$\lambda = \frac{h}{p}$

Where:

• $\lambda$ is the wavelength,


• $h$ is Planck's constant, which is approximately $6.626 \times 10^{-34} \, \text{Js}$ (joule-second),


• $p$ is the momentum of the object, calculated as the product of its mass ($m$) and velocity ($v$).

Given that the mass ($m$) of the ball is 100 g (which should be converted to kilograms to be consistent with the SI units), so $m = 0.1 \, \text{kg}$, and the velocity ($v$) of the ball is $100 \, \text{m/s}$, we can calculate the momentum ($p$) as:

$p = m \cdot v = 0.1 \, \text{kg} \cdot 100 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s}$

Now, substituting the values of $h$ and $p$ into the de Broglie wavelength equation:

$\lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{10 \, \text{kg} \cdot \text{m/s}}$

$\lambda = 6.626 \times 10^{-35} \, \text{m}$

So, the wavelength of the ball moving at a velocity of $100 \, \text{m/s}$ is $6.626 \times 10^{-35} \, \text{m}$, which is an extremely small wavelength, demonstrating the particle nature of macroscopic objects becomes significant only at the quantum scale.

(a) Probability of finding electrons at any place = $\psi _{2s}^2 = 0$ at node

$\therefore$ $\psi _{2s}^2 = 0 = {1 \over {4\sqrt {2\pi } }}{\left( {{1 \over {{a_0}}}} \right)^3}\left( {2 - {r \over {{a_0}}}} \right) \times {e^{ - r/{a_0}}}$

or $\left( {2 - {r \over {{a_0}}}} \right) = 0$ or $2 = {r \over {{a_0}}}$ or $r = 2{a_0}$

(b) $\lambda = {h \over {mv}}$

$h = 6.626 \times {10^{ - 34}}\,Js$, m = 100 g = 0.1 kg v = 100 ms^$-$1

$\therefore$ $\lambda = {{6.626 \times {{10}^{ - 34}}\,kg\,{m^2}\,{s^{ - 1}}} \over {0.1\,kg \times 100\,m{s^{ - 1}}}}$

$ = 6.626 \times {10^{ - 35}}\,m$

${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

$\therefore$ ${1 \over {91.2}} = {R_H}\left[ {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right]$

Since R_H = constant

${{{\lambda _{He}}} \over {{\lambda _H}}} = {{Z_H^2} \over {Z_{He}^2}} = {1 \over 4}$

or, ${\lambda _{He}} = {{{\lambda _H}} \over 4} = {{91.2} \over 4} = 22.8$ nm

No. of moles of H₂ gas,

$n = {{PV} \over {RT}} = {{1 \times 1} \over {0.0821 \times 298}} = 0.0409$

Energy required to dissociate 0.0409 mol H₂ into atoms $ = 0.0409 \times 436 = 17.83$ kJ

No. of moles of H atoms

$ = 2 \times 0.0409 = 0.0818$

${E_n} = - {{1312} \over {{n^2}}}$ kJ mol^$-$1

$\therefore$ Energy required for excitation of 0.0818 moles of H atoms from ground state to first excited state

$ = ({E_2} - {E_1}) \times 0.0818$ kJ mol^$-$1

$ = - \left[ {\left( {{{ - 1312} \over {{2^2}}}} \right) - \left( {{{ - 1312} \over {{1^2}}}} \right)} \right] \times 0.0818$ kJ

= 80.49 kJ

Total energy required for excitation of 1 litre H₂ gas = 17.83 + 80.49 = 98.32 kJ

To calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen, we need to understand what the Balmer series represents and how the wave number is related to the wavelength of electromagnetic radiation.

The Balmer series corresponds to electron transitions from higher energy levels (n > 2) down to n = 2 in the hydrogen atom. The wavelength of the emitted photon during such a transition is given by the Rydberg formula, which for hydrogen is:

$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$

where:

• $\lambda$ is the wavelength of the emitted light.
• $R$ is the Rydberg constant for hydrogen, approximately $1.097 \times 10^7 \, \text{m}^{-1}$.
• $n$ is the principal quantum number of the higher energy level (n > 2).

The wave number $\bar{\nu}$ is defined as the reciprocal of the wavelength (in meters), thus:

$\bar{\nu} = \frac{1}{\lambda}$

Substituting the Rydberg formula for $\frac{1}{\lambda}$ gives us:

$\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$

The shortest wavelength transition in the Balmer series corresponds to the transition with the largest energy difference, which occurs when the electron comes from an infinitely high energy level (n = ∞) down to n = 2. Therefore, for the shortest wavelength (and hence highest frequency and largest wave number), set n = ∞ in the equation. As $n$ approaches ∞, $\frac{1}{n^2}$ approaches 0, and the equation simplifies to:

$\bar{\nu}_{min} = R \left( \frac{1}{2^2} \right) = R \left( \frac{1}{4} \right) = \frac{1.097 \times 10^7 \, \text{m}^{-1}}{4}$

Performing the calculation:

$\bar{\nu}_{min} = 2.743 \times 10^6 \, \text{m}^{-1}$

Thus, the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen is $2.743 \times 10^6 \, \text{m}^{-1}$. This corresponds to the transition from n = ∞ to n = 2.

Bond energy per molecule $ = {{240 \times {{10}^3}\,J} \over {6.02 \times {{10}^{23}}}}$

$ = 3.98 \times {10^{ - 19}}$ J/molecule

Energy absorbed

$hc/\lambda = {{6.62 \times {{10}^{ - 34}}\,Js \times 3 \times {{10}^8}\,m\,{s^{ - 1}}} \over {4500 \times {{10}^{ - 10}}\,m}}$

$ = 4.41 \times {10^{ - 19}}\,J$

Kinetic energy

$ = (4.41 \times {10^{ - 19}} - 3.98 \times {10^{ - 19}})\,J$

$ = 4.3 \times {10^{ - 20}}\,J$

Kinetic energy per atom

$ = {{4.3 \times {{10}^{ - 20}}} \over 2}\,J = 2.15 \times {10^{ - 20}}\,J$

(i) The expression for the energy difference between two electronic levels is given by

$\overline v = {R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right){Z^2}$ ..... (ii)

$\Delta E = hc\,.\,{R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right){Z^2}$ ..... (iii)

Z = 1 for H atom

Therefore, energy difference between first and second orbit of H - atom is given by

$\Delta E = {R_H}\,.\,hc\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

$ = (1.09677 \times {10^7}\,{m^{ - 1}})(6.626 \times {10^{ - 34}}\,Js)(3 \times {10^8}\,m{s^{ - 1}}) \times \left( {{1 \over {{2^2}}} - {1 \over {{2^2}}}} \right)$

$ = 1.635 \times {10^{ - 18}}\,J$

(ii) Energy of X-rays of wavelength

$\lambda = {{hc} \over \lambda }$

Energy emission for hydrogen like atom during transition from n = 2 to n = 1 energy state = $1.635 \times {10^{ - 18}}\,{Z^2}J$

$\therefore$ $1.635 \times {10^{ - 18}}\,{Z^2}J = {{hc} \over \lambda }$

$ = {{(6.626 \times {{10}^{ - 34}}Js)(3 \times {{10}^8}\,m{s^{ - 1}})} \over {(3 \times {{10}^{ - 8}}\,m)}}$

$\therefore$ Z = 2

(iii) The hydrogen atom-like species having atomic number 2 is He⁺

For a spectral transition

${1 \over \lambda } = {R_H}{Z^2}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right)$

$\therefore$ For He⁺ ion, we have

${1 \over \lambda } = {R_H}\,.\,{(2)^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {3 \over 4}{R_H}$ ..... (i)

Now for hydrogen atom

${1 \over \lambda } = {R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right)$ ...... (ii)

Equating equations (i) and (ii), we get

$\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right) = {3 \over 4}$

Obviously n₁ = 1 and n₂ = 2. Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition, n = 4 to n = 2 in He⁺ species. The transition belongs to Lyman series.

To calculate the energy required to completely remove an electron from the n = 2 orbit in a hydrogen atom, we need to understand that completely removing an electron refers to ionizing the atom. Ionization requires the electron to move from its given orbit (in this case, n = 2) to essentially n = ∞, where its energy would be 0.

Given, the energy of an electron in a hydrogen atom is:

$ E = \frac{-21.7 \times 10^{-12}}{n^2} $ ergs.

For n = 2 (initial state energy $E_i$):

$ E_i = \frac{-21.7 \times 10^{-12}}{2^2} $

$ E_i = \frac{-21.7 \times 10^{-12}}{4} $

$ E_i = -5.425 \times 10^{-12} $ ergs.

For n = ∞ (final state energy, $E_f$), the energy is 0 since the electron is completely removed from the atom.

The energy required ($E_{required}$) to remove the electron is the difference in energy between the final and initial states:

$ E_{required} = E_f - E_i $

$ E_{required} = 0 - (-5.425 \times 10^{-12}) $

$ E_{required} = 5.425 \times 10^{-12} $ ergs.

To find the longest wavelength ($\lambda_{max}$) of light that can be used to cause this transition (which means providing exactly the energy calculated above), we use the relationship between energy and wavelength provided by the equation:

$ E = \frac{hc}{\lambda} $

Where:

• $E$ is the energy in ergs,


• $h$ is Planck's constant ($6.626 \times 10^{-27}$ erg·s),


• $c$ is the speed of light in vacuum ($3.00 \times 10^{10}$ cm/s),


• $\lambda$ is the wavelength in cm.

Rearranging this equation to solve for $\lambda$:

$ \lambda = \frac{hc}{E} $

Substituting the values:

$ \lambda_{max} = \frac{(6.626 \times 10^{-27}) \times (3.00 \times 10^{10})}{5.425 \times 10^{-12}} $

$ \lambda_{max} = \frac{1.9878 \times 10^{-16}}{5.425 \times 10^{-12}} $

$ \lambda_{max} = 3.664 \times 10^{-5} $ cm, or 366.4 nm.

Thus, the energy required to remove an electron completely from the n = 2 orbit is $5.425 \times 10^{-12}$ ergs, and the longest wavelength of light that can be used to cause this transition is $3.664 \times 10^{-5}$ cm (or 366.4 nm when converted to nanometers for more common wavelength units in light spectra considerations).

The outermost electronic configuration of an element refers to the arrangement of electrons in the outer shell or energy level closest to the nucleus that contains electrons. For chromium (Cr), which has the atomic number 24, its electron configuration can be initially thought to follow the typical filling order based on the Aufbau principle, which suggests an expected configuration of [Ar] 3d⁴ 4s², where [Ar] represents the electron configuration up to argon, a noble gas with a closed-shell configuration.

However, chromium is an exception to the typical filling order because half-filled and fully filled subshells are known to be more stable due to electron exchange energy and symmetry. As a result, one electron from the 4s orbital is used to fill the 3d orbital, giving chromium a more stable half-filled 3d subshell. Therefore, the actual outermost electron configuration of chromium is [Ar] 3d⁵ 4s¹.

This configuration emphasizes the stability of having a half-filled d subshell, which is energetically more favorable for chromium. This adjustment from the expected configuration demonstrates the importance of experimental observations in determining the actual electronic structures of elements, which sometimes deviate from theoretical predictions due to underlying quantum mechanical principles and electron-electron interactions.

The light radiations with discrete quantities of energy are called photons. Photons are elementary particles that are the quantum of the electromagnetic field, including electromagnetic radiation such as light, and the force carrier for the electromagnetic force. In quantum mechanics, the energy of a photon is quantized, meaning it exists only in discrete amounts. The energy $E$ of a photon is given by the equation $E = hf$, where $h$ is Planck's constant (approximately $6.626 \times 10^{-34} m^2 kg / s$) and $f$ is the frequency of the electromagnetic wave. This principle supports the particle theory of light, where light can be viewed as consisting of particles or photons, in contrast to the classical wave theory of light.

Wave functions of electrons in atoms and molecules are called orbitals. An orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The square of the orbital's magnitude |ψ|² represents the probability density of finding an electron at a particular point in space.