Join WhatsApp Icon JEE WhatsApp Group
Question 26

In a photoelectric effect experiment, the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be
Given E (in eV) = $$\frac{1237}{\lambda(\text{in nm})}$$

Step 1: Energy formula

$$E=\ \frac{\ 1237}{λ}$$

 Step 2: Work function (threshold energy)

$$ϕ=\ \frac{\ 1237}{380}​\approx3.26eV$$

 Step 3: Incident photon energy

$$E=\ \frac{\ 1237}{260}​\approx4.76eV$$

Step 4: Maximum kinetic energy

$$K_{\max}​=E−ϕ=4.76−3.26=1.50eV$$

Get AI Help

Video Solution

video

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests
Ask AI