To determine the nature of the graph of
B2 vs.
, we start by analyzing the given problem using the photoelectric effect and the behavior of electrons in a magnetic field.
Photoelectric Effect and Kinetic Energy
The energy of the incident light is given by:
E=where
h is Planck's constant,
c is the speed of light, and
λ is the wavelength of the incident light.
The kinetic energy (
K.E.) of the emitted photoelectrons, given the work function
φ, is:
K.E. =−φ Electron in a Magnetic Field
When these electrons enter a perpendicular magnetic field
B, they experience a force that causes them to move in a circular path. The centripetal force required for this motion is provided by the Lorentz force:
=evB where:
m is the mass of the electron,
v is the velocity of the electron,
R is the radius of the circular path,
e is the charge of the electron.
From the above equation, we can solve for
v :
v= Relationship between Kinetic Energy and Velocity
The kinetic energy of the electrons can also be expressed as:
K.E. =mv2Substituting
v= into the kinetic energy equation, we get:
m()2= K.E. = Combining Both Expressions for Kinetic Energy
We already have:
K.E. =−φEquating both expressions for kinetic energy:
=−φRearranging to isolate
B2 :
B2=(−φ)B2=⋅− Form of the Graph
This equation is in the form of a straight line
y=mx+c where:
y=B2x=m=c=− Thus, the graph of
B2 vs.
will be a straight line with a positive slope and a negative
y-intercept.
The correct graph will be a straight line with a positive slope and negative
y-intercept:
Option C