NEET + JEEAdvanced

Light of a single wavelength falls on two different metal surfaces.…

Question

Light of a single wavelength falls on two different metal surfaces. For a metal of threshold wavelength 540 nm the maximum kinetic energy of emitted photoelectrons is K1, and for a metal of threshold wavelength 600 nm it is K2, when both are illuminated by 360 nm light. Taking hc = 1240 eV·nm, find the value of (K2 - K1) in electron-volts.

✓ Verified answer: 0.2296checked by our engine — not a guess

Step-by-step solution

Photoelectric equation: K = hc/lambda - phi where phi = hc/lambda_th.
K1 = 1240/360 - 1240/540 = 3.4444 - 2.2963 = 1.1481 eV.
K2 = 1240/360 - 1240/600 = 3.4444 - 2.0667 = 1.3778 eV.
K2 - K1 = 1240/540 - 1240/600 = 2.2963 - 2.0667 = 0.2296 eV.

Final answer0.2296

Stuck on a problem like this?

Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.

Solve my doubt →

More Neet Modern Thermo solutions