Dual Nature of Radiation

According to Einstein's photoelectric equation,

$ \begin{aligned} & \left(E_K\right)_{\max }=E-\phi_0 \Rightarrow \phi_0=E-\left(E_K\right)_{\max } \\ & =8 \times 10^{-19}-\frac{p^2}{2 m} \\ & =8 \times 10^{-19}-\frac{\left(\frac{h}{\lambda}\right)^2}{2 m}=8 \times 10^{-19}-\frac{h^2}{2 m \lambda^2} \\ & =8 \times 10^{-19}-\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(10 \times 10^{-10}\right)^2} \\ & =8 \times 10^{-19}-2.41 \times 10^{-19} \\ & =5.59 \times 10^{-19} \mathrm{~J} \\ & =\frac{5.59 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=3.49 \mathrm{eV} \approx 3.5 \mathrm{eV} \end{aligned} $

$ \begin{aligned} & \lambda_{\min }=\frac{h c}{\mathrm{eV}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 20 \times 10^3} \\ & =6.2 \times 10^{-11} \mathrm{~m} \\ & =0.62 \times 10^{-10} \mathrm{~m}=0.62 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\begin{aligned} & \lambda=\frac{h}{\sqrt{2 \mathrm{meV}}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \frac{200}{3}}} \\ & =1.5 \times 10^{-10} \mathrm{~m}=1.5 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\lambda=\frac{h}{\sqrt{3 m k_B T}} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \lambda & \propto \frac{1}{\sqrt{T}} \\ \Rightarrow \quad \frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{(273+352)}{(273+127)}} \\ & =\sqrt{\frac{625}{400}}=\sqrt{\frac{25}{16}}=\frac{5}{4} \end{aligned} $

Step 1: Understanding Power

Power is how much energy is used or given out each second. So: $\text{Power} = \frac{\text{Energy}}{\text{Time}}$

Step 2: Energy of One Photon

The energy in one photon of light is found using: $E = h f$ where $h$ is Planck's constant and $f$ is the frequency.

Step 3: Total Energy from Many Photons

If $N$ photons come out in time $t$, the total energy is $N \times h f$.

Step 4: Linking Power and Number of Photons

So, $\text{Power} = \frac{\text{Total Energy}}{\text{Time}} = \frac{N \times h f}{t}$

Step 5: Average Number of Photons per Second

The number of photons per second is $n = \frac{N}{t} = \frac{P}{h f}$ Plug in the numbers: $n = \frac{33 \times 10^{-3}}{6.6 \times 10^{-34} \times 5 \times 10^{14}}$

Step 6: Final Calculation

This works out to: $n = 10 \times 10^{16}$ So, the laser gives off $10 \times 10^{16}$ photons every second.

Step 1: Formula for de-Broglie Wavelength

The de-Broglie wavelength is given by: $ \lambda = \frac{h}{\sqrt{2mk}} $

Step 2: Rearranging to Find Kinetic Energy

We can rearrange the formula to find kinetic energy $k$: $ \lambda^2 = \frac{h^2}{2mk} $ So, $ k = \frac{h^2}{2m\lambda^2} $

Step 3: Substitute the Values

Here,

• Planck's constant, $ h = 6.6 \times 10^{-34} $ Js
• Mass of electron, $ m = 9 \times 10^{-31} $ kg
• Wavelength, $ \lambda = 2 $ nm $ = 2 \times 10^{-9} $ m

Step 4: Calculation

Calculate the value: $ k = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times (2 \times 10^{-9})^2} $ This gives: $ k = 0.608 \times 10^{-19} \text{ J} $

Step 5: Convert Joules to Electron Volts

1 electron volt (eV) $ = 1.6 \times 10^{-19} $ J.
So, $ k = \frac{0.608 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} = 0.38 \text{ eV} $

The kinetic energy of the electron is about $ 0.38 $ eV.

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\begin{aligned} \lambda & =\frac{h}{\sqrt{2 m q V}} \\ \Rightarrow \quad \lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{m_\alpha}{m_p}} \sqrt{\frac{q_\alpha}{q_p}}=\sqrt{\frac{4}{1} \cdot \frac{2}{1}}=\frac{2 \sqrt{2}}{1} \\ \therefore \lambda_p: \lambda_\alpha & =2 \sqrt{2}: 1 \end{aligned} \end{aligned} $

Given:

Threshold wavelength,

$\lambda_0 = 6250 \, \text{Å} = 6250 \times 10^{-10} \, \text{m} = 6.25 \times 10^{-7} \, \text{m}$

Planck’s constant,

$h = 6.6 \times 10^{-34} \, \text{J·s}$

Speed of light,

$c = 3 \times 10^8 \, \text{m/s}$

Formula:

Work function $ \phi = h c / \lambda_0 $

Calculation:

$ \phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.25 \times 10^{-7}} $

$ \phi = \frac{19.8 \times 10^{-26}}{6.25 \times 10^{-7}} = 3.168 \times 10^{-19} \, \text{J} $

Converting to eV:

$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$

$ \phi = \frac{3.168 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.98 \, \text{eV} $

✅ Answer: Option B — 1.98 eV

$ \begin{aligned} & \text { Since, } E=\frac{h c}{\lambda} \\ & \begin{aligned} \Rightarrow \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}} \\ & =4.13 \times 10^{-7}=413 \times 10^{-9} \mathrm{~m} \\ & =413 \mathrm{~nm} \end{aligned} \end{aligned} $

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

when $\lambda^{\prime}=\lambda+0.25 \%$ of $\lambda=1.0025 \lambda$

So, $\quad 1.0025 \lambda=\frac{h}{p-p_0}$

$\Rightarrow \quad p-p_0=\frac{h}{1.0025}=\frac{p \lambda}{1.0025 \lambda}$

$\Rightarrow \quad \frac{p-p_0}{p}=\frac{1}{1.0025}$

$\Rightarrow \quad p=400 p_0$

$ \text { } \begin{aligned} E & =\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \\ = & 248 \times 10^{-9} \mathrm{~m}=248 \mathrm{~nm} \end{aligned} $

First, the photon energy is the sum of the work function and the maximum kinetic energy of the ejected electron:

• ϕ = 2 eV

• K_max = 2 eV

• E_photon = ϕ + K_max = 4 eV

Next, relate energy to wavelength via

$\lambda = \frac{hc}{E_{\rm photon}}$

with $hc = 1240\;\mathrm{eV\cdot nm}$. Therefore,

$\lambda = \frac{1240\;\mathrm{eV\cdot nm}}{4\;\mathrm{eV}} = 310\;\mathrm{nm} = 3100\;\text{\AA}.$

So the correct choice is Option B: 3100 Å.

Given:

Wavelength of incident light, $\lambda = 4000\, \text{Å} = 4 \times 10^{-7} \, \text{m}$

Threshold wavelength, $\lambda_0 = 5420\, \text{Å} = 5420 \times 10^{-10} \, \text{m}$

The relationship between the threshold wavelength $\lambda_0$ and the work function $\phi$ is:

$ \phi = \frac{hc}{\lambda_0} \, \text{J} \quad \Rightarrow \quad \phi = \frac{hc}{e \cdot \lambda_0} \, \text{eV} $

Calculating the work function:

$ \phi = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.602 \times 10^{-19} \times 5420 \times 10^{-10}} $

$ \phi = \frac{19.87 \times 10^3}{8682.84} = 2.288 \times 10^{-3} \times 10^3 $

$ \phi = 2.29 \, \text{eV} $

According to Einstein's photoelectric equation:

$ \frac{1}{2} mv^2 = \frac{hc}{\lambda} - \phi $

Where:

$ v $ is the velocity of the photoelectrons.

$ h $ is Planck's constant.

$ c $ is the speed of light.

$ \lambda $ is the wavelength of the incident light.

$ \phi $ is the work function of the metal.

Given the ratio of maximum velocities of photoelectrons $ v_1 = 3v_2 $, we analyze two cases:

Case 1:

$ \frac{1}{2} m (3v_2)^2 = \frac{hc}{\lambda_1} - \phi $

where $\lambda_1 = 300 \, \text{nm}$.

Case 2:

$ \frac{1}{2} m (v_2)^2 = \frac{hc}{\lambda_2} - \phi $

where $\lambda_2 = 500 \, \text{nm}$.

Dividing Equations for Case 1 and Case 2:

$ \frac{hc}{\lambda_1} - \phi = 9\left(\frac{hc}{\lambda_2} - \phi\right) $

Solving for $\phi$:

$ 8\phi = \frac{9hc}{\lambda_2} - \frac{hc}{\lambda_1} $

Substitute $ \frac{hc}{\lambda} $ with $ \frac{1240 \, \text{eV nm}}{\lambda} $:

$ 8\phi = \left(9 \times \frac{1240}{500} - \frac{1240}{300}\right) \, \text{eV} $

$ 8\phi = 18.18 \, \text{eV} $

Thus, solving for $\phi$:

$ \phi = \frac{18.18}{8} \approx 2.28 \, \text{eV} $

Therefore, the work function $\phi$ of the metal is approximately $2.28 \, \text{eV}$.

The maximum speed ratio of emitted photoelectrons due to different light wavelengths incident on a photosensitive metal surface is determined using:

$ \frac{v_1}{v_2} = \sqrt{\frac{\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_0}\right)}{\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_0}\right)}} $

Here,

$ v_1 $ is the speed of photoelectrons emitted due to light with wavelength $ \lambda_1 $.

$ v_2 $ is the speed of photoelectrons emitted due to light with wavelength $ \lambda_2 $.

Given:

$ \lambda_1 = \frac{\lambda_0}{3} $

$ \lambda_2 = \frac{\lambda_0}{9} $

The formula becomes:

$ \frac{v_1}{v_2} = \sqrt{\frac{\left(\frac{3}{\lambda_0} - \frac{1}{\lambda_0}\right)}{\left(\frac{9}{\lambda_0} - \frac{1}{\lambda_0}\right)}} $

Simplifying, we find:

$ \frac{v_1}{v_2} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2} $

Thus, the ratio of the maximum velocities of the photoelectrons is $ 1:2 $.

To find the number of photons emitted by a lamp, we can use Planck's quantum theory to calculate the energy of a photon. Given:

Mean wavelength ($\lambda$) = 4500 Å = $4500 \times 10^{-10}$ m

Planck's constant ($h$) = $6.626 \times 10^{-34}$ Js

Speed of light ($c$) = $3 \times 10^8$ m/s

Lamp power = 150 W

Efficiency = 8%

Energy of a Photon:

$ E = \frac{h c}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4500 \times 10^{-10}} $

$ E = 4.417 \times 10^{-19} \text{ J} $

Total Energy Emitted per Second:

Only 8% of the 150 W is emitted as light energy:

$ \text{Energy emitted as light} = \frac{8 \times 150}{100} = 12 \text{ J/s} $

Number of Photons Emitted:

Let $n$ be the number of photons emitted per second. The equation relating the total energy and the energy of photons is:

$ n \times \text{energy of a photon} = \text{total energy emitted as light} $

$ n \times 4.417 \times 10^{-19} = 12 $

Solve for $n$:

$ n = \frac{12}{4.417 \times 10^{-19}} $

$ n = 2.7167 \times 10^{19} $

$ n = 27167 \times 10^{18} $

Therefore, the number of photons emitted by the lamp per second is approximately $27.17 \times 10^{18}$.

The de Broglie wavelength is given by:

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

where $ p = \sqrt{2mK} $ represents the momentum, and $ K $ is the kinetic energy of the particle.

When the kinetic energy ($ K $) is decreased by 36%, the new energy ($ K' $) becomes:

$ K' = K - 0.36K = 0.64K $

The new momentum ($ p' $) can be calculated as follows:

$ p' = \sqrt{2mK'} = \sqrt{2m \cdot 0.64K} = 0.8p $

Given this new momentum, the new de Broglie wavelength ($ \lambda' $) is:

$ \lambda' = \frac{h}{p'} = \frac{h}{0.8p} = 1.25\lambda $

The change in wavelength ($ \Delta \lambda $) is:

$ \Delta \lambda = \lambda' - \lambda = 1.25\lambda - \lambda = 0.25\lambda $

The percentage increase in wavelength is:

$ \text{Percentage increase} = \frac{\Delta \lambda}{\lambda} \times 100 = 0.25 \times 100 = 25\% $

The de-Broglie wavelength is given by,

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

Initial de-Broglie wavelength is

$ \lambda=\frac{h}{m v_0} $

Equating the forces, we get

$ \Rightarrow \quad F=m a=e E_0 $

Acceleration of electron

$ \Rightarrow \quad a=\frac{e E_0}{m} $

Velocity of electron after time $t$

$ \Rightarrow \quad v=v_0+\frac{e E_0}{m} t $

So, de-Broglie wavelength after time $t$.

$ \begin{aligned} & \lambda^{\prime}=\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e E_0}{m} t\right)} \\ \Rightarrow & \frac{h}{m v_0\left(1+\frac{e E_0}{m v_0} t\right)}=\frac{\lambda}{\left(1+\frac{e E_0}{m v_0} t\right)} \end{aligned} $

To determine the longest wavelength of light that can cause the photoelectric effect in a metal with a work function of 9 eV, we use the formula:

$ \phi_0 = \frac{hc}{\lambda_0} $

Where:

$\phi_0$ is the work function (9 eV in this case).

$h$ is Planck's constant ($6.62 \times 10^{-34} \, \text{J s}$).

$c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$).

$\lambda_0$ is the wavelength.

Rearranging the formula to solve for $\lambda_0$, we get:

$ \lambda_0 = \frac{hc}{\phi_0} $

Substituting the given values:

$ \lambda_0 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{9 \times 1.6 \times 10^{-19}} $

Calculating this yields:

$ \lambda_0 = 1.37 \times 10^{-7} \, \text{m} $

Hence, the longest wavelength capable of initiating the photoelectric effect is $1.37 \times 10^{-7} \, \text{m}$.

The energy required to remove an electron from the aluminium surface is $ \phi_0 = 4.2 \, \text{eV} $.

The wavelength of the light is given as $ \lambda = 2000 \, \text{Å} = 2 \times 10^{-7} \, \text{m} $.

Using Einstein's photoelectric equation:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{h c}{\lambda} - \phi_0 $

Given the constants:

Planck's constant $ h = 6.6 \times 10^{-34} \, \text{J s} $

Speed of light $ c = 3 \times 10^8 \, \text{m/s} $

Conversion factor $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $

Mass of electron $ m = 9.1 \times 10^{-31} \, \text{kg} $

The calculation becomes:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7} \times 1.6 \times 10^{-19}} \, \text{eV} $

Further simplifying:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{6.2 \, \text{eV} - 4.2 \, \text{eV}}{9.1 \times 10^{-31}} $

$ v_{\text{max}}^2 = \frac{2 \times 2 \times 1.6 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31}} $

$ v_{\text{max}}^2 = \frac{6.4}{91} \times 10^{12} \quad \Rightarrow \quad v_{\text{max}} = \sqrt{\frac{6.4}{91} \times 10^{12}} $

Thus, the velocity of the fastest ejected electron is:

$ v_{\text{max}} = 0.838 \times 10^6 \quad \Rightarrow \quad v_{\text{max}} = 8.4 \times 10^5 \, \text{m/s} $

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)

Given, wavelength of incident light, $\lambda_1=1240 \mathop A\limits^o $

Stopping potential, $V_0=8 \mathrm{~V}$

Since, energy, $E=\frac{12400}{\lambda} \mathrm{eV}$

Incident energy $E_i=\frac{12400}{1240}=10 \mathrm{~eV}$

and $E_i-e V_0=W_0$

$\therefore W_0$ (work-function of emitter)

$\begin{aligned} & =(10-8) \mathrm{~eV}=2 \mathrm{~eV} \\ & \text { and } \\ & \lambda_0=\frac{12400}{2}=6200 ~\mathop A\limits^o \\ \end{aligned}$

Given, maximum speed of photoelectron $=v_{\max }$

According to Einstein's photoelectric equation,

$\begin{aligned} & \frac{1}{2} m v_{\max }^2=h \nu-h \nu_0 \\ & \Rightarrow \quad 2_{\max } \propto \nu \\ \end{aligned}$

So, graph between $v_{\max }$ and $v$ is a parabola as shown in option $(\mathrm{c})$.

Given, electric potential, $V=100 \mathrm{~V}$

Since, $\lambda=\frac{h}{p}=\frac{h}{m c}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}$

where, $\lambda$ is wavelength,

$h$ is Planck's constant i.e. $6.63 \times 10^{-34} \mathrm{Js}^{-1}$,

$m$ is mass of photon, $c$ is speed and $e$ is charge on proton

$\therefore \quad \lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times 200}}$

$=0.203\times10^{-11}$ m

$=2.03$ pm

It is near to 2.86 pm.

Given, wavelength of radiation $=300 \mathrm{~nm}$ $=300 \times 10^{-9} \mathrm{~m} $

Intensity $I=100 \mathrm{~W} / \mathrm{m}^2$

Number of photons falling on the surface,

$\begin{aligned} n_p & =\frac{\lambda}{h c}=\frac{300 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \\ & =0.1508 \times 10^{19} \end{aligned}$

As only $2 \%$ produce photoelectrons. So, number of photoelectrons ejected,

$n_e=\frac{2}{100} \times 0.1508 \times 10^{19}=0.302 \times 10^{17}$

$\therefore$ Number of photoelectrons emitted from $2 \mathrm{~cm}^2$

$\begin{aligned} \text { area } & =2 \times 10^{-4} \times n_e \\ & =0.604 \times 10^{13}=6.04 \times 10^{12} \end{aligned}$