Verified Solution • 2025–26 Session moderate

QJEEM21Ph9719: JEE Main 2021 Physics - Class 12 JEEM21Ph9719 Solution: Two stream of photons, possessing energies equal to twice and…

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ___________.

Step-by-Step Verified Solution

KEmax⁡=hv−ϕK{E_{\max }} = hv - \phi

12mv2=hv−ϕ{1 \over 2}m{v^2} = hv - \phi

v=2(hv−ϕ)mv = \sqrt {{{2(hv - \phi )} \over m}}

Given hv1=2ϕh{v_1} = 2\phi

hv2=10ϕh{v_2} = 10\phi

∴\therefore v1v2=hv1−ϕhv2−ϕ{{{v_1}} \over {{v_2}}} = \sqrt {{{h{v_1} - \phi } \over {h{v_2} - \phi }}}

v1v2=2ϕ−ϕ10ϕ−ϕ=13{{{v_1}} \over {{v_2}}} = \sqrt {{{2\phi - \phi } \over {10\phi - \phi }}} = {1 \over 3} = xy{x \over y}

∴\therefore x = 1

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring