Question

Let K₁ and K₂ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ₁ and λ₂, respectively are incident on a metallic surface. If λ₁ = 3λ₂ then:

• A. \(K_1>\frac{K_2}{3}\)
• B. \(K_1< \frac{K_2}{3}\)
• C. \(K_1=\frac{K_2}{3}\)
• D. \(K_2=\frac{K_1}{3}\)

Answer 1

The Correct Option is B

To solve this question, we need to apply the principles of the photoelectric effect as described by Albert Einstein in 1905. Let us understand the relationship between the wavelength of light and the maximum kinetic energy of photoelectrons emitted from a metallic surface:

1. The energy of a photon is given by \(E = \frac{hc}{\lambda}\), where \(h\) is the Planck's constant and \(c\) is the speed of light in vacuum.
2. According to the photoelectric effect, the maximum kinetic energy \(K\) of emitted photoelectrons is: \(K = \frac{hc}{\lambda} - \phi\) where \(\phi\) is the work function of the metal.
3. Given that the wavelengths are \(\lambda_1 = 3\lambda_2\), we can find the relationship between the kinetic energies \(K_1\) and \(K_2\).

For wavelength \(\lambda_1\): 

\(K_1 = \frac{hc}{\lambda_1} - \phi = \frac{hc}{3\lambda_2} - \phi\)

For wavelength \(\lambda_2\):

\(K_2 = \frac{hc}{\lambda_2} - \phi\)

1. From the above equations, we can equate the kinetic energies as follows:
2. \(K_1 = \frac{hc}{3\lambda_2} - \phi = \frac{1}{3} \left(\frac{hc}{\lambda_2} - \phi \right) + \frac{2}{3}\phi\)
3. Since \(\phi > 0\), therefore, \(\frac{2}{3}\phi > 0\), it implies that \(K_1 < \frac{1}{3}K_2\).

Thus, comparing the values, we get:

\(K_1 < \frac{K_2}{3}\)

Hence, the correct answer is: \(K_1 < \frac{K_2}{3}\).

Answer 2

The correct answer is (B) : \(K_1< \frac{K_2}{3}\)
\(K_1=\frac{hc}{λ_1}−\phi=\frac{hc}{3λ_2}−\phi….(i)\)
and
\(K_2=\frac{hc}{λ_2}−\phi….(ii)\)
from (i) and (ii) we can say
3K₁ = K₂ – 2φ
\(K_1<\frac{K_2}{3}\)