Home
Class 12
PHYSICS
Photons of energy 7eV fall on the surfac...

Photons of energy 7eV fall on the surface of a metal X resulting in emission of photoelectrons having maximum kinetic energy `E_(1) = 1eV`. Y is another metal on the surface of which photons of energy 6eV are incident and result in emission of photoelectrons of maximum kinetic energy `E_(2)= 2eV`. The ratio of work functions of metals X and Y, `(phi_(x))/(phi_(y))` is ____________

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the photoelectric equation, which relates the energy of the incident photons, the work function of the metal, and the maximum kinetic energy of the emitted photoelectrons. The equation is given by: \[ K_{\text{max}} = E - \phi \] where: - \( K_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons, - \( E \) is the energy of the incident photons, - \( \phi \) is the work function of the metal. ### Step 1: Analyze Metal X For metal X: - Energy of incident photons, \( E_X = 7 \, \text{eV} \) - Maximum kinetic energy of emitted photoelectrons, \( K_{1} = 1 \, \text{eV} \) Using the photoelectric equation: \[ K_{1} = E_X - \phi_X \] Substituting the known values: \[ 1 \, \text{eV} = 7 \, \text{eV} - \phi_X \] Rearranging gives: \[ \phi_X = 7 \, \text{eV} - 1 \, \text{eV} = 6 \, \text{eV} \] ### Step 2: Analyze Metal Y For metal Y: - Energy of incident photons, \( E_Y = 6 \, \text{eV} \) - Maximum kinetic energy of emitted photoelectrons, \( K_{2} = 2 \, \text{eV} \) Using the photoelectric equation: \[ K_{2} = E_Y - \phi_Y \] Substituting the known values: \[ 2 \, \text{eV} = 6 \, \text{eV} - \phi_Y \] Rearranging gives: \[ \phi_Y = 6 \, \text{eV} - 2 \, \text{eV} = 4 \, \text{eV} \] ### Step 3: Find the Ratio of Work Functions Now we have the work functions: - \( \phi_X = 6 \, \text{eV} \) - \( \phi_Y = 4 \, \text{eV} \) To find the ratio of the work functions: \[ \frac{\phi_X}{\phi_Y} = \frac{6 \, \text{eV}}{4 \, \text{eV}} = \frac{6}{4} = \frac{3}{2} \] ### Final Answer The ratio of the work functions of metals X and Y is: \[ \frac{\phi_X}{\phi_Y} = \frac{3}{2} \] ---

To solve the problem, we will use the photoelectric equation, which relates the energy of the incident photons, the work function of the metal, and the maximum kinetic energy of the emitted photoelectrons. The equation is given by: \[ K_{\text{max}} = E - \phi \] where: - \( K_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons, - \( E \) is the energy of the incident photons, - \( \phi \) is the work function of the metal. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV . The stopping voltage required for these electrons are

When a radiation of energy 5eV falls on a surface,the emitted photoelectrons have a maximum kinetic energy of 3eV .The stopping potential is

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6eV fall on it is 4eV. The stopping potential is:

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV . The stopping potential , in volt is

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

Violet light is falling on a photosensitive material causing ejection of photoelectrons with maximum kinetic energy of 1 eV . Red light falling on metal will cause emission of photoelectrons with maximum kinetic energy (approximately) equal to

A radiation of wavelength lamda illuminates a metal and ejects photoelectrons of maximum kinetic energy of 1eV. Aother radiation of wavelength (lamda)/(3) , ejects photoelectrons of maximum kinetic energy of 4eV. What will be the work function of metal?