Nature of radiation Questions in English

(A) The wavenumber $(\bar{\nu})$ is defined as the reciprocal of wavelength $(\lambda)$:
$\bar{\nu} = \frac{1}{\lambda}$
Given $\lambda = 5800 \ \mathring{A} = 5800 \times 10^{-8} \ cm = 5.8 \times 10^{-5} \ cm$.
$\bar{\nu} = \frac{1}{5.8 \times 10^{-5} \ cm} = 17241.37 \ cm^{-1}$.
We are given $\bar{\nu} = x \times 10 \ cm^{-1}$.
So,$x \times 10 = 17241.37$.
$x = 1724.137 \approx 1724$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 300 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \ J = 6.626 \times 10^{-19} \ J$.
Converting to $eV$ by dividing by $1.602 \times 10^{-19} \ J/eV$:
$E = \frac{6.626 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.136 \ eV$.
Photoelectric effect occurs if the work function $\phi < E$. Thus,metals with $\phi < 4.136 \ eV$ will show the effect.
Comparing the given values: $Li \ (2.4)$,$Na \ (2.3)$,$K \ (2.2)$,and $Mg \ (3.7)$ are all less than $4.136 \ eV$.
Therefore,there are $4$ such metals.

(A) The momentum of the photon is given by $p = \frac{h}{\lambda}$.
Substituting the values: $p = \frac{6.6 \times 10^{-34} \ J \ s}{330 \times 10^{-9} \ m} = 2 \times 10^{-27} \ kg \ m \ s^{-1}$.
According to the law of conservation of momentum,the momentum gained by the $He$ atom is equal to the momentum of the absorbed photon.
$p_{atom} = m \times v$,where $m$ is the mass of a single $He$ atom.
$m = \frac{\text{Molar mass}}{\text{Avogadro number}} = \frac{4 \times 10^{-3} \ kg \ mol^{-1}}{6 \times 10^{23} \ mol^{-1}} = \frac{2}{3} \times 10^{-26} \ kg$.
Equating the momenta: $2 \times 10^{-27} = (\frac{2}{3} \times 10^{-26}) \times v$.
$v = \frac{2 \times 10^{-27} \times 3}{2 \times 10^{-26}} = 0.3 \ m \ s^{-1}$.
Converting to $cm \ s^{-1}$: $0.3 \ m \ s^{-1} = 30 \ cm \ s^{-1}$.

(C) According to Moseley's law,the frequency $\nu$ of the characteristic $X$-rays emitted by an element is related to its atomic number $Z$ by the equation: $\sqrt{\nu} = a(Z - b)$,where $a$ and $b$ are constants.
This implies that a plot of $\sqrt{\nu}$ versus atomic number $(Z)$ is a straight line.
Statement $(I)$ is false because the plot of $\sqrt{\nu}$ versus atomic mass is not a straight line.
Statement $(II)$ is false because the plot of $\nu$ (not $\sqrt{\nu}$) versus atomic number is not a straight line; rather,$\sqrt{\nu}$ versus atomic number is a straight line.
Therefore,both statements are false.

(A) The momentum $P$ of a photon is given by the relation $P = \frac{E}{c},$ where $E$ is the energy and $c$ is the speed of light.
Given $E = 1 \ MeV = 1 \times 10^6 \ eV = 1 \times 10^6 \times 1.602 \times 10^{-19} \ J = 1.602 \times 10^{-13} \ J.$
Speed of light $c = 3 \times 10^8 \ ms^{-1}.$
$P = \frac{1.602 \times 10^{-13} \ J}{3 \times 10^8 \ ms^{-1}} = 0.534 \times 10^{-21} \ kg \ ms^{-1}.$

(C) $\gamma$-rays are electromagnetic radiations with zero charge and zero rest mass.
Due to their electromagnetic nature,they travel with the speed of light,which is approximately $3 \times 10^8 \ m/s$.

(B) Frequency is defined as the number of waves that pass through a given point in one second.
It is measured in Hertz $(Hz)$,where $1 \ Hz = 1 \ s^{-1}$.
- $(A)$ Wavelength: The distance between two consecutive crests or troughs.
- $(C)$ Wave number: The number of wavelengths per unit length $(1 / \lambda)$.
- $(D)$ Amplitude: The maximum displacement of the wave from its mean position.

(A) The energy of one photon is given by $E = \frac{hc}{\lambda}$.
Given: $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m/s$,$\lambda = 700 \ nm = 700 \times 10^{-9} \ m$.
Energy of one photon $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} \ J = 2.84 \times 10^{-19} \ J$.
To find the energy per mole,multiply by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
$E_{mole} = E \times N_A = 2.84 \times 10^{-19} \times 6.022 \times 10^{23} \ J/mol \approx 1.71 \times 10^5 \ J/mol$.

(B) The relationship between frequency $(\nu)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Given:
$c = 3 \times 10^8 \ ms^{-1}$
$\lambda = 600 \ nm = 600 \times 10^{-9} \ m = 6 \times 10^{-7} \ m$.
Substituting the values into the formula:
$\nu = \frac{3 \times 10^8 \ ms^{-1}}{6 \times 10^{-7} \ m} = 0.5 \times 10^{15} \ Hz = 5.0 \times 10^{14} \ Hz$.
Therefore,the correct option is $B$.

(C) The relationship between wavelength $(\lambda)$ and wave number $(\bar{\nu})$ is given by: $\lambda = \frac{1}{\bar{\nu}}$
Given $\bar{\nu} = 11516 \ cm^{-1}$.
$\lambda = \frac{1}{11516 \ cm^{-1}} \approx 8.6835 \times 10^{-5} \ cm$.
To convert this to nanometers $(nm)$,we use the conversion factor $1 \ nm = 10^{-7} \ cm$:
$\lambda = 8.6835 \times 10^{-5} \ cm \times \frac{1 \ nm}{10^{-7} \ cm} = 868.35 \ nm \approx 868 \ nm$.

(B) The wavenumber $(\bar{\nu})$ is defined as the reciprocal of the wavelength $(\lambda)$.
$\bar{\nu} = \frac{1}{\lambda}$
Given,$\lambda = 0.25 \mu m = 0.25 \times 10^{-6} \ m$.
Substituting the value in the formula:
$\bar{\nu} = \frac{1}{0.25 \times 10^{-6} \ m} = 4.0 \times 10^6 \ m^{-1}$.

(D) The $Balmer$ series refers to a series of emission spectral lines in the visible region of the electromagnetic spectrum for hydrogen.
These lines are produced when an electron transitions from a higher energy level $(n \geq 3)$ to the second energy level $(n=2)$ of the hydrogen atom.
The wavelengths of the lines in the $Balmer$ series fall within the visible light spectrum.
They correspond to the colors observed in hydrogen's emission spectrum,including red,blue,and violet lines.

(B) When an electron jumps from any higher energy level $(n_2 > 1)$ to the ground state $(n_1 = 1)$,the resulting emission spectral lines belong to the Lyman series.
Since the transition is from $n_2 = \infty$ to $n_1 = 1$,it represents the limiting line of the Lyman series.

(D) The energy of a photon is given by the equation $E = \frac{hc}{\lambda}$.
This shows that energy is inversely proportional to the wavelength $(\lambda)$.
Shorter the wavelength,higher is the frequency,and higher is the energy.
Among the given colours,violet light has the shortest wavelength (approximately $400 \ nm$),therefore it possesses the highest energy.

(B) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m \ s^{-1}$ and $\lambda = 750 \ nm = 750 \times 10^{-9} \ m$.
Substituting the values: $v = \frac{3 \times 10^8 \ m \ s^{-1}}{750 \times 10^{-9} \ m} = \frac{3}{750} \times 10^{17} \ Hz = 0.004 \times 10^{17} \ Hz = 4 \times 10^{14} \ Hz$.

(B) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\lambda = 460 \ nm = 460 \times 10^{-9} \ m$.
Substituting the values: $v = \frac{3 \times 10^8 \ m/s}{460 \times 10^{-9} \ m} = 6.52 \times 10^{14} \ Hz$.
Thus,the frequency is approximately $6.5 \times 10^{14} \ Hz$.

(C) The relationship between speed of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is given by $c = \nu \times \lambda$.
Given: $c = 3 \times 10^8 \ m/s$ and $\nu = 50 \ Hz$.
Rearranging the formula for wavelength: $\lambda = \frac{c}{\nu}$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{50} = 0.06 \times 10^8 \ m = 6 \times 10^6 \ m$.

(C) The frequency $v$ is given by the formula $v = \frac{c}{\lambda}$.
Given wavelength $\lambda = 480 \ nm = 480 \times 10^{-9} \ m$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting the values: $v = \frac{3 \times 10^8 \ m/s}{480 \times 10^{-9} \ m} = 6.25 \times 10^{14} \ Hz$.

(C) The frequency $\nu$ is calculated using the formula $\nu = \frac{c}{\lambda}$.
Given: Speed of light $c = 3 \times 10^8 \ m \ s^{-1}$ and wavelength $\lambda = 400 \ nm = 400 \times 10^{-9} \ m$.
Substituting the values: $\nu = \frac{3 \times 10^8 \ m \ s^{-1}}{400 \times 10^{-9} \ m} = 7.5 \times 10^{14} \ Hz$.

(D) The photoelectric effect occurs when light of a suitable frequency strikes the surface of a metal,causing the ejection of electrons.
The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,provided the frequency is above the threshold frequency.
The kinetic energy of the emitted electrons depends on the frequency of the incident light,not its intensity.
Therefore,the correct statement is that the number of electrons ejected increases with the increase in the intensity of incident light.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Since $h$ and $c$ are constants,$E \propto \frac{1}{\lambda}$.
Therefore,the ratio of energies is inversely proportional to the ratio of their wavelengths:
$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$
Given $\frac{E_1}{E_2} = \frac{3}{2}$,we have $\frac{3}{2} = \frac{\lambda_2}{\lambda_1}$.
Thus,the ratio of wavelengths $\lambda_1 : \lambda_2 = 2:3$.

(B) The deflection of a charged particle in a magnetic field is given by the Lorentz force,$F = qvB \sin \theta$,and the resulting acceleration is $a = F/m = (qvB \sin \theta)/m$.
Since the deflection is proportional to the charge-to-mass ratio $(q/m)$,the particle with the highest $q/m$ ratio will undergo the maximum deflection.
$\alpha$-particles $(He^{2+})$ have a mass of approximately $4 \ amu$ and a charge of $+2$.
$\beta$-particles $(e^-)$ have a mass of approximately $1/1837 \ amu$ and a charge of $-1$.
$\gamma$-rays are electromagnetic radiation with no charge and no mass,so they show no deflection.
Neutrons have no charge,so they show no deflection.
Comparing the $q/m$ ratios,the $\beta$-particle has a significantly higher $q/m$ ratio than the $\alpha$-particle,leading to maximum deflection.

(A) Statement $I$ is incorrect because isotopes of an element have the same electronic configuration and therefore exhibit similar chemical properties.
Statement $II$ is correct because the Lyman series corresponds to electronic transitions to the $n=1$ energy level,which releases energy in the ultraviolet $(UV)$ region.
Statement $III$ is correct because,according to Maxwell's wave theory,electromagnetic radiation consists of oscillating electric and magnetic fields that are mutually perpendicular to each other and to the direction of wave propagation.
Therefore,statements $II$ and $III$ are correct.

(A) The energy of electromagnetic radiation is given by $E = h\nu = \frac{hc}{\lambda}$.
For $a$: $\nu_a = 3 \times 10^{15} \ Hz$.
For $b$: $\nu_b = 2 \times 10^{14} \ Hz$.
For $c$: $\lambda_c = 400 \ nm = 400 \times 10^{-9} \ m$. Frequency $\nu_c = \frac{c}{\lambda_c} = \frac{3 \times 10^8}{400 \times 10^{-9}} = 7.5 \times 10^{14} \ Hz$.
For $d$: $\lambda_d = 750 \ nm = 750 \times 10^{-9} \ m$. Frequency $\nu_d = \frac{c}{\lambda_d} = \frac{3 \times 10^8}{750 \times 10^{-9}} = 4 \times 10^{14} \ Hz$.
Comparing frequencies: $\nu_b (2 \times 10^{14}) < \nu_d (4 \times 10^{14}) < \nu_c (7.5 \times 10^{14}) < \nu_a (30 \times 10^{14})$.
Since $E \propto \nu$,the increasing order of energies is $b < d < c < a$.

(A) The relationship between wavelength $(\lambda)$,frequency $(\nu)$,and the speed of light $(c)$ is given by the formula: $\lambda = \frac{c}{\nu}$.
Given: $c = 3 \times 10^8 \ m \ s^{-1}$ and $\nu = 103.4 \ MHz = 103.4 \times 10^6 \ s^{-1} = 1.034 \times 10^8 \ s^{-1}$.
Substituting the values: $\lambda = \frac{3 \times 10^8 \ m \ s^{-1}}{1.034 \times 10^8 \ s^{-1}} \cong 2.90 \ m$.

(C) The spectral series of hydrogen are defined by the lower energy level $n_2$ to which the electron transitions from higher levels $n_1$:
$n_2 = 1$ (Lyman series): Ultraviolet region.
$n_2 = 2$ (Balmer series): Visible region.
$n_2 = 3$ (Paschen series): Infrared region.
$n_2 = 4$ (Brackett series): Infrared region.
$n_2 = 5$ (Pfund series): Infrared region.
Since the question refers to transitions to $n = 3, 4, 5$,all these series (Paschen,Brackett,and Pfund) fall within the Infrared region.

(D) The wavelength $\lambda$ is given as $7.8 \times 10^2 \ nm$.
Convert the wavelength to meters: $\lambda = 7.8 \times 10^2 \times 10^{-9} \ m = 7.8 \times 10^{-7} \ m$.
The frequency $\nu$ is calculated using the formula $\nu = \frac{c}{\lambda}$,where $c = 3 \times 10^8 \ m \ s^{-1}$ is the speed of light.
$\nu = \frac{3 \times 10^8 \ m \ s^{-1}}{7.8 \times 10^{-7} \ m} \approx 3.846 \times 10^{14} \ s^{-1}$.
Thus,the frequency is approximately $3.8 \times 10^{14} \ s^{-1}$.

(C) The energy of the incident radiation $(E)$ is calculated using the formula: $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \text{ eV} \approx 3.1 \text{ eV}$.
For photoemission to occur,the energy of the incident radiation must be greater than or equal to the work function $(E \ge W_0)$.
Conversely,electrons will not be ejected if the work function is greater than the incident energy $(W_0 > E)$.
Comparing the given work functions with $3.1 \text{ eV}$:
$Mg (3.7 > 3.1)$,$Cu (4.8 > 3.1)$,and $Ag (4.3 > 3.1)$ have work functions greater than $3.1 \text{ eV}$.
Therefore,$3$ metals will not eject electrons.

(D) According to Einstein's photoelectric equation,the kinetic energy $(KE)$ of the ejected photoelectron is given by: $KE = h\nu - \varphi$,where $h\nu$ is the energy of incident radiation and $\varphi$ is the work function of the metal.
From this relation,it is clear that for a constant incident energy $(h\nu)$,the kinetic energy $(KE)$ is inversely proportional to the work function $(\varphi)$ of the metal.
Therefore,the metal with the highest work function will result in the lowest kinetic energy of the ejected photoelectrons.
The order of work functions for the given metals is: $Cu (4.70 \ eV) > Ag (4.26 \ eV) > Li (2.90 \ eV) > Na (2.75 \ eV)$.
Since $Cu$ has the highest work function,it will exhibit the lowest kinetic energy for the ejected photoelectrons.

(C) Given wavelength $\lambda = 6300 \ \mathring{A} = 6300 \times 10^{-10} \ m = 6.3 \times 10^{-7} \ m$.
$(I)$ Wave number $(\bar{\nu})$ is the reciprocal of wavelength:
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.3 \times 10^{-7} \ m} \approx 1.587 \times 10^6 \ m^{-1}$.
$(II)$ Frequency $(\nu)$ is given by $\nu = \frac{c}{\lambda}$,where $c = 3 \times 10^8 \ m \ s^{-1}$:
$\nu = \frac{3 \times 10^8 \ m \ s^{-1}}{6.3 \times 10^{-7} \ m} \approx 4.761 \times 10^{14} \ s^{-1}$.
Thus,the correct option is $C$.

(C) The kinetic energy $(KE)$ of photoelectrons is given by the equation: $KE = h\nu - h\nu_0$.
When the incident frequency $\nu > \nu_0$,then $KE > 0$. Thus,statement $3$ is correct.
Here,$\nu$ is the frequency of incident light and $\nu_0$ is the threshold frequency.
Analyzing other statements:
$1$. The number of photoelectrons ejected is directly proportional to the intensity of light,not inversely.
$2$. When $\nu < \nu_0$,$KE < 0$,which is physically impossible; therefore,the photoelectric effect cannot be observed.
$4$. When $\nu = \nu_0$,$KE = 0$,meaning the electrons are just ejected but have no kinetic energy to move away from the surface.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 450 \ nm = 450 \times 10^{-9} \ m$.
$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{450 \times 10^{-9} \ m} \approx 4.417 \times 10^{-19} \ J$.
Converting to $eV$: $E = \frac{4.417 \times 10^{-19} \ J}{1.602 \times 10^{-19} \ J/eV} \approx 2.76 \ eV$.
Photoelectric effect occurs if the energy of the incident photon is greater than or equal to the work function $(E \ge W_0)$.
Metals with $W_0 < 2.76 \ eV$ are $K (2.25 \ eV), Na (2.30 \ eV)$,and $Li (2.42 \ eV)$. These will undergo the photoelectric effect.
Metals with $W_0 > 2.76 \ eV$ are $Mg (3.70 \ eV)$ and $Cu (4.80 \ eV)$. These will not undergo the photoelectric effect.
The number of such metals is $2$.

(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 300 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.626 \times 10^{-19} \ J$.
Converting this energy to $eV$: $E = \frac{6.626 \times 10^{-19}}{1.6022 \times 10^{-19}} \approx 4.135 \ eV$.
Photoelectric effect occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.
Comparing $\Phi$ values:
$Ag: 4.3 \ eV > 4.135 \ eV$ (No emission)
$Mg: 3.7 \ eV < 4.135 \ eV$ (Emission)
$K: 2.25 \ eV < 4.135 \ eV$ (Emission)
$Na: 2.30 \ eV < 4.135 \ eV$ (Emission)
Thus,$Mg, K,$ and $Na$ will eject electrons. The total number of such metals is $3$.

(D) The energy $(E)$ of electromagnetic radiation is inversely proportional to its wavelength $(\lambda)$ according to the equation $E = \frac{hc}{\lambda}$.
As the wavelength increases,the energy of the radiation decreases.
The electromagnetic spectrum in order of decreasing energy is: $\gamma-rays > X-rays > UV > visible > IR > microwave > radio \ waves$.
Comparing this with the given options,the correct order is $X-rays > UV > IR > microwave$.

(C) The hydrogen atomic spectrum consists of five series of spectral lines,which fall into specific regions of the electromagnetic spectrum:
$1$. Lyman series: $UV$ region
$2$. Balmer series: Visible region
$3$. Paschen series: $IR$ region
$4$. Brackett series: $IR$ region
$5$. Pfund series: $IR$ region
Comparing these with the given options:
$(i)$ $\gamma$-radiation: Not observed.
$(ii)$ $UV$: Observed (Lyman series).
$(iii)$ $X$-rays: Not observed.
$(iv)$ Infrared: Observed (Paschen,Brackett,Pfund series).
Therefore,$\gamma$-radiation and $X$-rays are not seen in the hydrogen atomic spectrum.

(A) Based on the electromagnetic spectrum,the frequency ranges are as follows:
| Name | Frequency (in $Hz$) |
| :--- | :--- |
| $X$-rays | $10^{17}$ to $10^{21}$ |
| Radiowaves | $10^{5}$ to $10^{8}$ |
| $UV$ rays | $10^{15}$ to $10^{16}$ |
| $IR$ rays | $10^{12}$ to $10^{13}$ |
Comparing the given ranges,$X$-rays have the highest frequency range ($10^{17}$ to $10^{21} \ Hz$).
Therefore,the correct option is $A$.

(D) The energy of radiation is given by the formula: $E = h \nu = \frac{hc}{\lambda} = hc \bar{\nu}$,where $\bar{\nu}$ is the wave number.
Rearranging for the wave number: $\bar{\nu} = \frac{E}{hc}$.
Substituting the given values: $\bar{\nu} = \frac{19.875 \times 10^{-13} \ erg}{6.625 \times 10^{-27} \ erg \ sec \times 3 \times 10^{10} \ cm \ sec^{-1}}$.
$\bar{\nu} = \frac{19.875 \times 10^{-13}}{19.875 \times 10^{-17}} = 10^4 \ cm^{-1}$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 310 \ nm = 310 \times 10^{-9} \ m$.
Using $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$,$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{310 \times 10^{-9}} \approx 6.41 \times 10^{-19} \ J$.
Converting to $eV$: $E = \frac{6.41 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.0 \ eV$.
Photoelectric effect occurs if the energy of the incident photon is greater than or equal to the work function $(\Phi)$ of the metal. If $E < \Phi$,the photoelectric effect does not occur.
Here,$E = 4.0 \ eV$.
Comparing with work functions:
$M_1: 4.8 \ eV > 4.0 \ eV$ (No effect)
$M_2: 4.3 \ eV > 4.0 \ eV$ (No effect)
$M_3: 4.75 \ eV > 4.0 \ eV$ (No effect)
$M_4: 3.75 \ eV < 4.0 \ eV$ (Effect occurs)
Thus,$M_1, M_2$ and $M_3$ do not show the photoelectric effect.

(D) The energy of the incident radiation is calculated using the formula $E = \frac{hc}{\lambda}$.
For $\lambda = 300 \ nm = 300 \times 10^{-9} \ m$,$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{300 \times 10^{-9} \ m} \approx 6.626 \times 10^{-19} \ J \approx 4.14 \ eV$.
The work functions $(\Phi)$ of the metals are approximately: $Li \approx 2.5 \ eV$,$Mg \approx 3.7 \ eV$,$Ag \approx 4.3 \ eV$,$Cu \approx 4.7 \ eV$,and $K \approx 2.3 \ eV$.
Photoelectric effect occurs when $E > \Phi$.
Comparing the values,$Li, Mg,$ and $K$ have work functions less than $4.14 \ eV$.
Thus,$3$ metals show the photoelectric effect.

(B) Statement $(i)$ is correct: The photoelectric effect is an instantaneous process with no time lag between light incidence and electron emission.
Statement $(ii)$ is incorrect: The number of photoelectrons ejected is directly proportional to the intensity of the incident light.
Statement $(iii)$ is correct: Alkali metals like $K$,$Rb$,and $Cs$ have low work functions and exhibit the photoelectric effect even with visible light.
Therefore,statements $(i)$ and $(iii)$ are correct.

(C) The energy of the photon is given as $E = 3 \ eV$.
Converting energy to $erg$: $E = 3 \times 1.6 \times 10^{-12} \ erg = 4.8 \times 10^{-12} \ erg$.
Using the formula $\lambda = \frac{hc}{E}$:
$\lambda = \frac{6.626 \times 10^{-27} \ erg \ s \times 3 \times 10^{10} \ cm/s}{4.8 \times 10^{-12} \ erg}$.
$\lambda = \frac{19.878 \times 10^{-17}}{4.8 \times 10^{-12}} \ cm = 4.141 \times 10^{-5} \ cm$.
Since $1 \ cm = 10^8 \ \mathring{A}$,then $\lambda = 4.141 \times 10^{-5} \times 10^8 \ \mathring{A} = 4141 \ \mathring{A}$.

(D) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Substituting the values:
$v = \frac{3 \times 10^8 \ m/s}{600 \times 10^{-9} \ m} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} \ Hz = 5.0 \times 10^{14} \ Hz$.

(C) The energy required to break one mole of $H_2$ bonds is $436 \ kJ \ mol^{-1}$.
To break a single bond,the energy required $(E)$ is $\frac{436 \times 10^3 \ J \ mol^{-1}}{6.023 \times 10^{23} \ mol^{-1}} \approx 7.24 \times 10^{-19} \ J$.
The relationship between energy and wavelength is $E = \frac{h \cdot c}{\lambda}$,where $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m \cdot s^{-1}$.
Rearranging for wavelength: $\lambda = \frac{h \cdot c}{E}$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m \cdot s^{-1}}{7.24 \times 10^{-19} \ J} \approx 2.745 \times 10^{-7} \ m$.
Converting to nanometers: $\lambda \approx 274.5 \ nm \approx 274 \ nm$.

(A) The energy of a photon is given by $E = h \nu = \frac{hc}{\lambda} = hc \cdot \bar{\nu}$,where $\bar{\nu} = \frac{1}{\lambda}$.
For option $(A)$: $\lambda = 300 \ nm = 300 \times 10^{-9} \ m$. $E = \frac{(6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{8} \ m/s)}{300 \times 10^{-9} \ m} = 6.626 \times 10^{-19} \ J$.
For option $(B)$: $\nu = 3 \times 10^{4} \ s^{-1}$. $E = h \nu = (6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{4} \ s^{-1}) = 1.9878 \times 10^{-29} \ J$.
For option $(C)$: $\bar{\nu} = 30 \ cm^{-1} = 3000 \ m^{-1}$. $E = hc \cdot \bar{\nu} = (6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{8} \ m/s)(3000 \ m^{-1}) = 5.9634 \times 10^{-22} \ J$.
For option $(D)$: $E = 6.626 \times 10^{-27} \ J$.
Comparing the values,the energy in option $(A)$ is the highest.

(B) $(1)$. Wavelength of $A = 400 \ nm = 400 \times 10^{-9} \ m$.
$(2)$. Wavelength of $B$ $(\lambda) = \frac{c}{\nu} = \frac{3 \times 10^{8} \ m/s}{10^{16} \ s^{-1}} = 3 \times 10^{-8} \ m = 30 \ nm$.
$(3)$. Wavelength of $C$ $(\lambda) = \frac{1}{\bar{\nu}} = \frac{1}{10^{4} \ cm^{-1}} = 10^{-4} \ cm = 10^{-6} \ m = 1000 \ nm$.
Comparing wavelengths: $\lambda_{C} (1000 \ nm) > \lambda_{A} (400 \ nm) > \lambda_{B} (30 \ nm)$.
Since Energy $(E) = \frac{hc}{\lambda}$,$E \propto \frac{1}{\lambda}$.
Therefore,the order of energy is $E_{B} > E_{A} > E_{C}$.

(C) The energy $E$ of a radiation is inversely proportional to its wavelength $\lambda$,given by the formula $E = \frac{hc}{\lambda}$.
Therefore,the ratio of energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 2000 \mathring{A}$ and $\lambda_2 = 6000 \mathring{A}$.
Substituting the values,we get $\frac{E_1}{E_2} = \frac{6000}{2000} = 3$.