Exams › JEE Advanced › Physics

In a photoelectric effect experiment, the incident wavelength is lambda. When the wavelength is reduced by (2/3)*lambda (i.e., the new wavelength is lambda/3), the maximum kinetic energy of the emitted photoelectrons becomes n times the original value. What is the threshold wavelength of the metal?

1. ((n-1)/(n-3))*lambda
2. (n/(n-3))*lambda
3. ((n+1)*lambda)/(n-3)
4. 3*lambda/n

Correct answer: ((n-1)/(n-3))*lambda

Solution

Let K = initial max KE, lambda0 = threshold wavelength. K = hc/lambda - hc/lambda0... (1). n*K = 3*hc/lambda - hc/lambda0... (2). Subtract (1) from (2): (n-1)*K = 2*hc/lambda => K = 2*hc/[(n-1)*lambda]. From (1): hc/lambda0 = hc/lambda - K = hc/lambda - 2*hc/[(n-1)*lambda] = hc/lambda * [1 - 2/(n-1)] = hc/lambda * (n-3)/(n-1). So lambda0 = lambda*(n-1)/(n-3).

Related JEE Advanced Physics questions

• The photoelectric effect provides evidence for the quantum behavior of light because:
• A metal surface is exposed to light with wavelengths of 248 nm and 310 nm. The maximum velocities of the emitted photoelectrons for these wavelengths are v₁ and v₂, respectively. Given that the ratio v₁: v₂ is 2: 1 and hc = 1240 eV·nm, estimate the work function of the metal.
• An electron in a higher energy level of a Li²⁺ ion possesses an angular momentum of 3h/2π. The electron's de Broglie wavelength in this energy state is expressed as pπa₀ (where a₀ represents the Bohr radius). Determine the value of p.
• In an experiment to calculate Planck's constant, light with varying wavelengths was shone on a metal surface, and the energy of the emitted photoelectrons was determined using stopping potentials. The following data was recorded for the wavelength (λ) of the light and the corresponding stopping potential (V₀): λ (μm) | V₀ (V) 0.3 | 2.0 0.4 | 1.0 0.5 | 0.4 Using c = 3 × 10⁸ m/s and e = 1.6 × 10⁻¹⁹ C, the value of Planck's constant (in J·s) obtained from this experiment is:
• A photoelectric material having work-function ϕ₀ is illuminated with light of wavelength λ (λ < hc/ϕ₀). The fastest photoelectron has a de Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then the ratio Δλd/Δλ is proportional to:
• In a photoelectric experiment, metals P, Q, and R emit photoelectrons with maximum kinetic energies EP, EQ, and ER, respectively, such that EP = 2EQ = 2ER. The same monochromatic light source is used for metals P and Q, while a different monochromatic light source is used for metal R. The work functions of metals P, Q, and R are 4.0 eV, 4.5 eV, and 5.5 eV, respectively. What is the energy of the photons incident on metal R, in eV?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →