($h = 6.63 \times 10^{-34} \text{ J.s}, e = 1.6 \times 10^{-19} \text{ C}, c = 3 \times 10^8 \text{ m/s}$)
Photoelectric Effect Analysis
The photoelectric effect relates photon energy ($E_{photon}$) to the metal's work function ($\phi$) and the maximum kinetic energy ($KE_{max}$) of emitted electrons using Einstein's equation: $E_{photon} = \phi + KE_{max}$.
Expressing energy in terms of wavelength ($\lambda$) and stopping potential ($V_s$), we have $E_{photon} = \frac{hc}{\lambda}$ and $KE_{max} = eV_s$. The equation becomes $\frac{hc}{\lambda} = \phi + eV_s$.
Formulating Equations
We are given:
- Light 1: Wavelength = $\lambda_1$, Stopping Potential $V_{s1} = 3.2 \text{ V}$.
- Light 2: Wavelength = $\lambda_2 = 2\lambda_1$, Stopping Potential $V_{s2} = 0.7 \text{ V}$.
Applying the photoelectric equation for both cases:
- For Light 1: $ \frac{hc}{\lambda_1} = \phi + e(3.2) \quad \quad (1) $
- For Light 2: $ \frac{hc}{2\lambda_1} = \phi + e(0.7) \quad \quad (2) $
Solving for $\lambda_1$
Subtract equation (2) from equation (1) to eliminate the work function ($\phi$):
$ (\frac{hc}{\lambda_1} - \frac{hc}{2\lambda_1}) = (\phi + 3.2e) - (\phi + 0.7e) $
$ \frac{hc}{2\lambda_1} = (3.2 - 0.7)e $
$ \frac{hc}{2\lambda_1} = 2.5e $
Rearranging to solve for $\lambda_1$:
$ \lambda_1 = \frac{hc}{5e} $
Calculation
Substitute the given physical constants: $h = 6.63 \times 10^{-34} \text{ J.s}$, $c = 3 \times 10^8 \text{ m/s}$, and $e = 1.6 \times 10^{-19} \text{ C}$:
$ \lambda_1 = \frac{(6.63 \times 10^{-34} \text{ J.s}) \times (3 \times 10^8 \text{ m/s})}{5 \times (1.6 \times 10^{-19} \text{ C})} $
$ \lambda_1 = \frac{19.89 \times 10^{-26} \text{ J.m}}{8 \times 10^{-19} \text{ C}} $
$ \lambda_1 = 2.48625 \times 10^{-7} \text{ m} $
The calculation yields $\lambda_1 \approx 2.49 \times 10^{-7}$ m. The provided correct option is B.
Similar Questions
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
The binding energy for the following nuclear reactions are expressed in MeV.
${}_2\text{He}^3 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^4 + 20 \text{ MeV}$
${}_2\text{He}^4 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^5 - 0.9 \text{ MeV}$
If $\text{X}_3, \text{X}_4, \text{X}_5$ denote the stability of ${}_2\text{He}^3, {}_2\text{He}^4$ and ${}_2\text{He}^5$, respectively, then the correct order is :Identify the correct truth table of the given logic circuit.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{\text{nd}}$ and $4^{\text{th}}$ orbits of hydrogen atom __________.
- The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - The energy of an electron in an orbit of the Bohr's atom is $-0.04 E_0 \text{ eV}$ where $E_0$ is the ground state energy. If $L$ is the angular momentum of the electron in this orbit and $h$ is the Planck's constant, then $\frac{2\pi L}{h}$ is __________ :
- A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$) Truth table for the following digital circuit will be :
Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
The binding energy for the following nuclear reactions are expressed in MeV.
${}_2\text{He}^3 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^4 + 20 \text{ MeV}$
${}_2\text{He}^4 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^5 - 0.9 \text{ MeV}$
If $\text{X}_3, \text{X}_4, \text{X}_5$ denote the stability of ${}_2\text{He}^3, {}_2\text{He}^4$ and ${}_2\text{He}^5$, respectively, then the correct order is :Identify the correct truth table of the given logic circuit.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$)
