Question

Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV.The ratio of maximum speed of emitted electrons is

A. 1:4
B. 1:5
C. 1:3
D. 1:2

Answer 1

The ratio of maximum speed of emitted electrons is 1:2.

So the correct answer is option D). 1:2.

Step-by-step explanation:

To find the ratio of maximum speed of emitted electrons when light of two different frequencies illuminates a metal with a given work function, one can use the photoelectric equation:

E(photon) = K.E.(max) + Work Function

For the first photon with an energy of 1eV:

1 eV = K.E.(max1) + 0.5 eV

K.E.(max1) = 1 eV - 0.5 eV

= 0.5 eV

From this kinetic energy, one can use the kinetic energy equation:

K.E.(max) = (1/2)m*v(max)^2

To solve for v(max), the maximum speed given the kinetic energy.

The mass (m) of the electron is a constant value, so the kinetic energy is proportional to the square of maximum speed.

For the second photon with an energy of 2.5eV:

2.5 eV = K.E.(max2) + 0.5 eV

K.E.(max2) = 2 eV

This is again converted to maximum speed using the kinetic energy equation.

Now to find the ratio of the two maximum speeds, v(max1)/v(max2), we can use the ratios of the square root of their kinetic energies, since K.E. is proportional to v^2:

v(max1):v(max2) = √(K.E.(max1)):√(K.E.(max2))

v(max1):v(max2) = √(0.5 eV):√(2 eV)

This simplifies to:

v(max1):v(max2) = 1:√4

v(max1):v(max2) = 1:2.