Question:
\(K_1\) and \(K_2\) be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength \(\lambda_1\) and \(\lambda_2\), respectively. If \(\lambda_1 = 2\lambda_2\) then the work function of material is given by:
\(K_1\) and \(K_2\) be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength \(\lambda_1\) and \(\lambda_2\), respectively. If \(\lambda_1 = 2\lambda_2\) then the work function of material is given by:
Updated On: Apr 13, 2026
- \(K_2 + 2K_1\)
- \(2K_2 - K_1\)
- \(K_1 - 2K_2\)
- \(K_2 - 2K_1\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Einstein's photoelectric equation states that the maximum kinetic energy of an emitted photoelectron is the difference between the energy of the incident photon and the work function ($\phi$) of the metal.
Step 2: Key Formula or Approach:
1. $K = \frac{hc}{\lambda} - \phi$.
2. Two equations: $K_1 = \frac{hc}{\lambda_1} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$.
Step 3: Detailed Explanation:
Given $\lambda_1 = 2\lambda_2$. From equation 1: $K_1 = \frac{hc}{2\lambda_2} - \phi \implies 2(K_1 + \phi) = \frac{hc}{\lambda_2}$. From equation 2: $K_2 = \frac{hc}{\lambda_2} - \phi \implies K_2 + \phi = \frac{hc}{\lambda_2}$. Equating the two expressions for $\frac{hc}{\lambda_2}$: \[ 2K_1 + 2\phi = K_2 + \phi \] \[ 2\phi - \phi = K_2 - 2K_1 \] \[ \phi = K_2 - 2K_1 \]
Step 4: Final Answer:
The work function is $K_2 - 2K_1$.
Einstein's photoelectric equation states that the maximum kinetic energy of an emitted photoelectron is the difference between the energy of the incident photon and the work function ($\phi$) of the metal.
Step 2: Key Formula or Approach:
1. $K = \frac{hc}{\lambda} - \phi$.
2. Two equations: $K_1 = \frac{hc}{\lambda_1} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$.
Step 3: Detailed Explanation:
Given $\lambda_1 = 2\lambda_2$. From equation 1: $K_1 = \frac{hc}{2\lambda_2} - \phi \implies 2(K_1 + \phi) = \frac{hc}{\lambda_2}$. From equation 2: $K_2 = \frac{hc}{\lambda_2} - \phi \implies K_2 + \phi = \frac{hc}{\lambda_2}$. Equating the two expressions for $\frac{hc}{\lambda_2}$: \[ 2K_1 + 2\phi = K_2 + \phi \] \[ 2\phi - \phi = K_2 - 2K_1 \] \[ \phi = K_2 - 2K_1 \]
Step 4: Final Answer:
The work function is $K_2 - 2K_1$.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main Photoelectric Effect Questions
- In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and the stopping potential is 2V, what is the wavelength of the electromagnetic wave? Given \( hc = 1242 \, \text{eV} \cdot \text{nm} \) where \( h \) is the Planck constant and \( c \) is the speed of light in vacuum.
- A light source of wavelength \( \lambda \) illuminates a metal surface, and electrons are ejected with a maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength \( \frac{\lambda}{2} \), then the maximum kinetic energy of ejected electrons will be (The work function of the metal is 1 eV).
- The work function of a metal 3eV. The colour of the visible light that is required to cause emission of photo electrons is:
- For a certain metal, when monochromatic light of wavelength \(\lambda\) is incident, the stopping potential for photoelectrons is \(3V_0\). When the same metal is illuminated by light of wavelength \(2\lambda\), the stopping potential becomes \(V_0\). The threshold wavelength for photoelectric emission for the given metal is \(\alpha \lambda\). The value of \(\alpha\) is:
- The graph shows variation of stopping potential \(V_0\) with the frequency \(\nu\) of the incident radiation for three photosensitive metals \(X_1\), \(X_2\) and \(X_3\). Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions