Dual Nature of Radiation

$3.8 = 0.6 + {1 \over 2}mv_1^2$

$1.4 = 0.6 + {1 \over 2}mv_2^2$

$ \Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}$

$ \Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}$

Given, $v_{\max }=1.6 \times 10^6 \mathrm{~m} / \mathrm{s}$

$ e=1.6 \times 10^{-19} \mathrm{C}, m=9 \times 10^{-31} \mathrm{~kg} $

If $V_0$ be the stopping potential, then according to Einstein's photoelectric equation,

$ \begin{aligned} e V_0 & =\frac{1}{2} m v_{\max }^2 \\ \Rightarrow \quad V_0 & =\frac{m v_{\max }^2}{2 e} \\ & =\frac{9 \times 10^{-31} \times\left(1.6 \times 10^6\right)^2}{2 \times 1.6 \times 10^{-19}}=7.2 \mathrm{~V} \end{aligned} $

Photon flux = No. of photon reaching surface per unit area per unit time

$ \begin{aligned} & =\frac{\text { No. of photon emitted per sec }}{\text { Area of incidence }} \\ & =\frac{P}{E_1 \times A}=\frac{P}{\frac{h c}{\lambda} \times A} \\ & =\frac{942 \times 660 \times 10^{-9}}{\left(6.6 \times 10^{-34} \times 3 \times 10^8 \times 4 \times \pi \times 5^2\right)} \\ & \approx 1 \times 10^{19} \end{aligned} $

By increasing the potential difference between cathode and anode continuously in a photoelectric experiment, the photocurrent increases only upto certain value and after that remains constant (saturate).

Hence, statement I is false.

According to given situation is statement II,

$ \begin{aligned} \frac{\left(K_{\max }\right)_A}{\left(K_{\max }\right)_B} & =\frac{E_A-\phi}{E_B-\phi} \\ & =\frac{2.5-2}{3.5-2}=\frac{0.5}{1.5}=\frac{1}{3} \end{aligned} $

Hence, statement II is also false.

The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal surface.

Hence, statement III is also false.

de-Broglie wavelength of a moving particle is given as

$ \begin{array}{ll} \lambda =\frac{h}{m v} \\ \Rightarrow \quad \lambda \propto \frac{1}{m v} \end{array} $

Since, product of $m v$ for a ball of mass 0.03 kg moving with velocity $1 \mathrm{~m} / \mathrm{s} =1 \times 0.03=0.03 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ which is minimum among the all given options. Therefore, de-Broglie wavelength of ball of mass 0.03 kg moving with velocity $1 \mathrm{~m} / \mathrm{s}$ is largest.

According to Einstein's photoelectric equation,

$ e V_0=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) $

when, $\lambda=\lambda_1=200 \mathrm{~nm}$, then

$ e\left(V_0\right)_1=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_0}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

when, $\lambda=\lambda_2=400 \mathrm{~nm}$

$ \therefore e\left(V_0\right)_2=h c\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_0}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∴ Subtracting Eq. (ii) form Eq. (i),

$ \begin{aligned} & e\left(V_0\right)_1-e\left(V_0\right)_2=h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_0}-\frac{1}{\lambda_2}+\frac{1}{\lambda_0}\right] \\ & \begin{aligned} \Rightarrow e\left[\left(V_0\right)_1-\left(V_0\right)_2\right] & =h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \\ e\left[\left(V_0\right)_1-\left(V_0\right)_2\right] & =1240 \mathrm{eV} \\ - & \mathrm{nm}\left[\frac{1}{200 \mathrm{~nm}}-\frac{1}{400 \mathrm{~nm}}\right] \\ \Rightarrow\left(V_0\right)_1-\left(V_0\right)_2 & =1240\left[\frac{2-1}{400}\right] \\ = & 1240 \times \frac{1}{400}=3.1 \mathrm{~V} \end{aligned} \end{aligned} $

Given, kinetic energy of electron,

$ \begin{aligned} K & =320 \mathrm{eV} \\ & =320 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =512 \times 10^{-17} \mathrm{~J} \\ h & =6 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned} $

Mass of electron,

$ \begin{aligned} m_e & =9 \times 10^{-31} \mathrm{~kg} \\ e & =1.6 \times 10^{-19} \mathrm{C} \end{aligned} $

∴ de-Broglie wavelength of electron,

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 \mathrm{mK}}} \\ & =\frac{6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 512 \times 10^{-17}}} \\ & =\frac{6 \times 10^{-34}}{9.6 \times 10^{-24}} \\ & =0.625 \times 10^{-10} \mathrm{~m} \\ & =62.5 \times 10^{-12} \mathrm{~m} \\ & =62.5 \mathrm{pm}\end{aligned} $

$ \quad\left[\because 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right] $

Given, wavelength of incident radiation, $\lambda=248 \mathrm{~nm}=248 \times 10^{-7} \mathrm{~m}$

Stopping potential, $V_0=2.8 \mathrm{eV}$ According to Einstein's equation,

$ \begin{aligned} e V_0 & =h v-\phi_0 \\ \Rightarrow \quad e V_0 & =\frac{h c}{\lambda}-\phi_0 \\ \Rightarrow \quad \phi_0 & =\frac{h c}{\lambda}-e V_0 \\ & =\frac{1240}{248}-2.8=5-2.8 \\ & =2.2 \mathrm{eV} \end{aligned} $

Given, potential difference,

$ \begin{aligned} & V=121 \mathrm{~V} \\ & h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}, m=9 \times 10^{-31} \mathrm{~kg} \end{aligned} $

de-Broglie wavelength is given as

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 \mathrm{meV}}} \\ & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 121}} \\ & =0.112 \times 10^{-9} \mathrm{~m} \\ & =0.112 \mathrm{~nm} \end{aligned} $

Given, slope of graph between stopping potential versus frequency is equal to $\frac{h}{e}$.

where, $h$ is planck's constant.

$ \begin{array}{ll} \therefore \frac{h}{e}=\text { slope }=4 \times 10^{-15} \\ \Rightarrow \quad h=e \times 4 \times 10^{-15} \\ =1.6 \times 10^{-19} \times 4 \times 10^{-15}=6.4 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{array} $

In photoelectric effect, kinetic energy of emitted photoelectrons is directly proportional to the frequency of emitted photoelectrons. Hence, on increasing frequency of incident radiation, kinetic energy of emitted photoelectron increases. Hence, statement I is incorrect and statement IV is correct.

In photoelectric effect, photoelectric emission takes place only when frequency of incident radiation is greater than or equal to threshold frequency, no matter how much intensity of incident light is. Hence, statement II is not correct.

In photoelectric emission, maximum kinetic energy of emitted photoelectrons depends only frequency of incident radiation and does not depend on intensity of incident radiation. Hence, statement III is correct.

Photon energy is dependent on the frequency or wavelength of radiation. It does not depend on the intensity of radiation.

Since, $E=h v=\frac{h c}{\lambda}$

i.e. $E \propto v$ or $E \propto \frac{1}{\lambda}$

where, $\mathrm{v}=$ frequency and

$\lambda=$ wavelength.

Given, transition levels,

$ n_1=2, n_2=3 $

Energy emitted during transition

$ n_2=3 \text { to } n_1=2 \text { in H-atom } $

(i.e. for $\mathrm{H}_\alpha$ - light)

$ \begin{aligned} E & =13.6\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ & =13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =13.6\left[\frac{5}{36}\right] \mathrm{eV}=1.89 \mathrm{eV} \end{aligned} $

For $\mathrm{H}_\alpha$-light to be able to emit photoelectrons from a metal, the work function must be less than or equal to 1.89 eV .

Hence, maximum value of work function $=1.89 \mathrm{eV}$

The correct statement regarding the photoelectric effect is:

The maximum kinetic energy of emitted photoelectrons in photoelectric emission depends on the frequency of the incident radiation and is independent of the intensity of the incident radiation. Therefore, the statement given in option (c) is correct. For a given frequency of incident radiation, the photocurrent varies with changes in the intensity of the incident radiation. Additionally, for a given frequency of radiation, the stopping potential does not depend on the intensity of the incident radiation.

It is also important to note that for a frequency lower than the cutoff frequency, photoelectric emission cannot occur regardless of how much the intensity of the incident radiation is increased.

The Wavelength of incident photon, $\lambda=800 \mathrm{~nm}$ The wavelength of emitted photoelectron, $\lambda_1=1 \mathrm{~nm}$ Since, wavelength of emitted photoelectron is less then that of incident photon, it means the photoelectrons gains energy higher than the incident proton. Which is not possible in photoelectric emission. Hence, the work function of the metal cannot be calculated with the given data.

Work function of given metals are given below

$\begin{aligned} & \phi_{\mathrm{Na}}=2.0 \mathrm{~eV} \\ & \phi_{\mathrm{Al}}=4.2 \mathrm{~eV} \\ & \phi_{\mathrm{Pt}}=5.65 \mathrm{~eV} \\ & \phi_{\mathrm{Cs}}=2.14 \mathrm{~eV} \end{aligned}$

Hence, work function for platinum is highest (5.65 eV)