Dual Nature of Radiation and Matter

A. $E=h v$ is energy of photon

B. Diffraction and interference confirm wave nature of light

C. $\lambda=\frac{h}{p}$ is de Broglie wavelength of particle.

D. Compton effect confirms particle nature of light.

For incident radiation having wavelength ( $\lambda$ ), photoelectric effect doesn't occur when $\frac{h c}{\lambda}<$ work-function

$ \begin{aligned} & \Rightarrow \lambda>\frac{h c}{W_0} \\ & \Rightarrow \lambda>\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}} \\ & \Rightarrow \lambda>\frac{3 \times 10^{-7}}{1.6} \\ & \Rightarrow \lambda>\frac{300}{1.6} \mathrm{~nm} \\ & \Rightarrow \lambda=187.5 \mathrm{~nm} \end{aligned} $

∴ Option (3) 200 nm is correct.

$ \begin{aligned} & E_e=E_{P h}=20 \mathrm{eV} \\ & \Rightarrow P_e=\sqrt{2 m_e E_e}, P_{p h}=\frac{E_{p h}}{c} \\ & \Rightarrow \frac{P_e}{P_{P h}}=\frac{\sqrt{2 m_e E_e}}{E_{P h}} c=\sqrt{\frac{2 m_e}{E_{P h}} c} \\ & \frac{P_e}{P_{P h}}=\sqrt{\frac{2 \times 9 \times 10^{-31}}{20 \times 1.6 \times 10^{-19}}} \times 3 \times 10^8 \\ & =\frac{3}{4} \times 10^{-6} \times 3 \times 10^8=\frac{900}{4}=225 \end{aligned} $

Power of source is decreasing that means number of photon is decreasing linearly with time and source is same so, energy of photon is same.

Power $(P) \propto n E$

$P \propto n$ (number of photons per second)

$ \text { and energy of photon is constant. } $

For ejection of electron energy of incident photon should be greater than work function

$ \begin{aligned} & \phi=\frac{h c}{\lambda} \\ & \phi_1=\frac{h c}{\lambda_1} \\ & \lambda_1<\lambda \\ & \phi_1>\phi \end{aligned} $

No electron will eject

$ V_1=0 $

By Einstein photoelectric equation

$ \begin{aligned} & \phi=\phi_0+e V \\ & \lambda_2>\lambda \\ & \lambda_3 \gg \lambda \\ & \phi>\phi_2 \\ & \phi \gg \phi_3 \end{aligned} $

So, $V_3>V_2$

The De-Broglie wavelength of an electron in the $n=2$ state of a hydrogen atom can be calculated as follows:

Firstly, calculate the radius $r$ for the $n=2$ state using the Bohr radius formula:

$ r = 0.052 n^2 $

For $n=2$, the radius is:

$ \begin{aligned} r &= 0.052 \times 4 \\ &= 0.208 \, \text{nm} \end{aligned} $

We use the relationship $M v r = \frac{n h}{2 \pi}$, leading to the expression for the De-Broglie wavelength $\lambda$:

$ \lambda = \frac{h}{M v} = \pi r $

Substituting the calculated $r$ value:

$ \begin{aligned} \lambda &= \pi \times 0.208 \, \text{nm} \\ &= 3.14 \times 0.208 \, \text{nm} \\ &= 0.65317 \, \text{nm} \\ &\approx 0.67 \, \text{nm} \end{aligned} $

Thus, the De-Broglie wavelength of the electron in the $n=2$ state is approximately $0.67$ nm.

Photoelectric current is directly proportional to intensity of light.

To find the ratio of the de Broglie wavelengths of a photon and an electron when they both have the same energy $E$, follow these steps:

Photon Wavelength:

For a photon, the energy is given by:

$ E = \frac{h c}{\lambda_{\mathrm{Ph}}} $

Where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_{\mathrm{Ph}}$ is the wavelength of the photon. Rearranging this equation gives:

$ \lambda_{\mathrm{Ph}} = \frac{h c}{E} $

Electron Wavelength:

For an electron, the energy related to its momentum is:

$ E = \frac{p^2}{2m} $

Where $m$ is the mass of the electron and $p$ is its momentum. Using the de Broglie wavelength expression $p = \frac{h}{\lambda_e}$, we can write:

$ E = \left( \frac{h}{\lambda_e} \right)^2 \times \frac{1}{2m} $

Solving for the electron's wavelength $\lambda_e$:

$ \lambda_e = \frac{h}{\sqrt{2mE}} $

Ratio of Wavelengths:

Now, calculate the ratio of the wavelengths $\frac{\lambda_{\mathrm{Ph}}}{\lambda_e}$ as follows:

$ \frac{\lambda_{\mathrm{Ph}}}{\lambda_e} = \frac{\frac{h c}{E}}{\frac{h}{\sqrt{2mE}}} = c \sqrt{\frac{2m}{E}} $

This ratio equation defines the relation between the photon and electron wavelengths when both have the same energy $E$.

de-Broglie wavelength is given by

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}$

For same potential difference

$\begin{gathered} \lambda \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}} \\ \because m_\alpha \gg m_e \\ \lambda_e>\lambda_\alpha \end{gathered}$

The energy of the incident light is given by:

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

where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. Substituting the given value of $\lambda$, we get:

$E = \frac{hc}{hc/e} = e$

The maximum kinetic energy of the photoelectron is given by:

$KE_{max} = E - \phi$

where $\phi$ is the work function. Substituting the values, we get:

$KE_{max} = e - \phi$

Therefore, the correct answer is Option C.

de-Broglie wavelength $\lambda=\frac{h}{P}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$ where $E=\frac{1}{2} m v^2$

Squaring both sides,

$\begin{aligned} & \lambda^2=\frac{h^2}{4 m^2 E} \\ & \Rightarrow \frac{1}{\lambda^2}=\text { (constant) } E \end{aligned}$

Graph passes through origin with constant slope.

A. The energy of a photon is $E = h v$.

This statement is correct. The energy of a photon is given by the equation:

$E = h v$

where $E$ is the energy, $h$ is Planck's constant, and $v$ (also written as $\nu$) is the frequency of the photon.

B. The velocity of a photon is $c$.

This statement is also correct. In free space (vacuum), the velocity of light, and hence the velocity of a photon, is constant and is denoted by $c$.

C. The momentum of a photon, $p = \frac{h v}{c}$.

This statement is correct as well. The momentum $p$ of a photon can be expressed as:

$p = \frac{E}{c} = \frac{h v}{c}$

D. In a photon-electron collision, both total energy and total momentum are conserved.

This statement is correct. During collisions involving photons and electrons, such as Compton scattering, both energy and momentum are conserved.

E. Photon possesses positive charge.

This statement is incorrect. Photons are electrically neutral particles and do not possess any charge.

Based on the analysis, the correct statements are A, B, C, and D. Therefore, the correct answer is:

Option B
A, B, C, and D only

${\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o = {{12.27} \over {\sqrt {81} }}\mathop A\limits^o = {{12.27} \over 9}\mathop A\limits^o $

$ = 1.36\mathop A\limits^o $ ($\because$ $1\mathop A\limits^o = {1 \over {10}}$ nm)

$ = 0.136$ nm

Maximum kinetic energy of emitted electron is independent of intensity of radiation.

Given energy of photon $E=2.20 ~\mathrm{eV}$

Work function of $\mathrm{Cs} ~\phi_{0}=2.14 ~\mathrm{eV}, \mathrm{K} ~\phi_{0}=2.30 ~\mathrm{eV}, \mathrm{Na} ~\phi_{0}=2.75 ~\mathrm{eV}$

We know that $e^{-}$ emitts when $h v>\phi_{0}$

here it is clear that energy of photon is more than the work function of $\mathrm{Cs}$ [Caesium] only so Ans. only (Cs).

The minimum wavelength, $\lambda_{\min}$, of X-rays produced by an electron can be determined using the equation derived from the energy of a photon and the energy given to an electron by an accelerating voltage, $V$.

The energy of a photon is given by $E = hc/\lambda$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the photon.

When an electron is accelerated through a potential difference of $V$ volts, it gains kinetic energy equal to $eV$, where $e$ is the electron charge.

This energy is then converted into a photon's energy when the electron collides with a target in an X-ray tube, resulting in X-rays of wavelength $\lambda$. Setting the kinetic energy equal to the photon energy gives $eV = hc/\lambda$. Solving for $\lambda$ gives $\lambda = hc/(eV)$.

Therefore, the minimum wavelength (corresponding to the maximum energy photon produced when all the kinetic energy is converted into photon energy) is inversely proportional to the voltage, $V$.

Thus, $\lambda_{\min} \propto 1/V$.

We know, $K{E_{\max }} = hv - h{v_0}$

$e{V_0} = hv - h{v_0}$ ($\because$ $K{E_{\max }} = e{V_0}$)

$e{V_0} = 4.2\,eV - 2.2\,eV$

$\therefore$ ${V_0} = 2\,V$

According to Einstein's photoelectric equation

${(K.E)_{\max }} = hv - h{v_0}$

${(K.E)_{\max }} = h(4{v_0}) - h{v_0}$

${(K.E)_{\max }} = 3h{v_0}$

Since ${k_{\max }} = e{V_s} = hv - \phi $

${{e{V_s}} \over 2} = hv - h{v_0}$ ....... (i)

$e{V_s} = {{hv} \over 2} - h{v_0}$ ...... (ii)

${1 \over 2}\left[ {{{hv} \over 2} - h{v_0}} \right] = hv - h{v_0}$

$ \Rightarrow h{v_0} - {{h{v_0}} \over 2} = hv - {{hv} \over 4}$

$ \Rightarrow {{h{v_0}} \over 2} = {{3hv} \over 4}$

${v_0} = {{3v} \over 2}$

* Language of question is wrongly framed. The values of stopping potentials should be interchanged.

Given, temperature of sun, $T_s=6000 \mathrm{~K}$

Radius of sun, $R_s=7.2 \times 10^5 \mathrm{~km}=7.2 \times 10^8 \mathrm{~m}$

Radius of earth, $R_e=6000 \mathrm{~km}=600 \times 10^3 \mathrm{~m}$

Distance between earth and sun,

$d=15 \times 10^7 \mathrm{~km}=15 \times 10^{10} \mathrm{~m}$

Intensity of light on the earth,

$\begin{aligned} I & =\frac{\text { Total energy emitted by sun }}{4 \pi d^2} \times\left(\pi R_e{ }^2\right) \\ & =\frac{\sigma\left(T_s^4 \cdot 4 \pi R_s^2 \times \pi R_e{ }^2\right)}{4 \pi d^2}=\frac{\sigma T_s^4 R_s{ }^2 \times \pi R_e{ }^2}{d^2} \\ & \begin{aligned} & =\frac{5.67 \times 10^{-8} \times(6000)^4 \times\left(7.2 \times 10^8\right)^2 \times 3.14 \times (6000\times10^3)^2}{\left(15 \times 10^{10}\right)^2} \\ & =19.2 \times 10^{16} \end{aligned} \end{aligned}$

In radioactivity, electrons are emitted in the form of $\beta\left({ }_{-1} e^{\circ}\right)$ particles. It is also called negative $\beta$-decay, where an unstable nucleus emits an energetic electron and an anti-neutrino and a neutron in the nucleus becomes a proton that remains in the product nucleus.

In photoelectric effect, when light rays of frequency greater than or equal to threshold frequency incident on metal surface, then electrons are emitted from metal surface. This phenomenon is called photoelectric effect. For photoelectric emission it is not necessary that element is unstable but for radioactivity, it is necessary that elements are unstable. Hence, assertion is true but reason is false.

As we know that, wavelength of a particle

$\begin{aligned} & \lambda =\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m}} \cdot \frac{1}{\sqrt{E}} \\ \Rightarrow \quad & \log \lambda =\log \left(\frac{h}{\sqrt{2 m}} \cdot \frac{1}{\sqrt{E}}\right) \\ \Rightarrow \quad & \log \lambda =\log \frac{h}{\sqrt{2 m}}+\log \frac{1}{E^{1 / 2}} \\ \Rightarrow \quad & \log \lambda =\log \left(\frac{h}{\sqrt{2 m}}\right)-\frac{1}{2} \log E \\ \Rightarrow \quad & \log \lambda =-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2 m}} \end{aligned}$

So, equation of straight line is $y=m x+c$.

This is equation of line with slope $-\frac{1}{2}$.

The photoelectric effect is the emission of electrons (called photoelectrons) when light strikes a metal surface. Hence, photoelectric effect can take place only with an electron bound in the atom. Electron is a lepton whereas proton is fermion.