Verified Solution • 2025–26 Session moderate

QJEEM19Ph78dc: JEE Main 2019 Physics - Class 12 JEEM19Ph78dc Solution: Surface of certain metal is first illuminated with light of…

Surface of certain metal is first illuminated with light of wavelength λ\lambda 1 = 350 nm and then, by light of wavelength λ\lambda 2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

(Energy of photon n = 1240λ(in mm){{1240} \over {\lambda (in\,mm)}} eV)

Multiple Choice Options

  • A 1.8
  • B 2.5
  • C 5.6
  • D 1.4

Step-by-Step Verified Solution

Let speed of photon electron in first case is 2v

then in the second case speed is v.

For first case

hcλ1=ϕ+12{{hc} \over {{\lambda _1}}} = \phi + {1 \over 2} m(2v)2

For second case,

hcλ2=ϕ+12{{hc} \over {{\lambda _2}}} = \phi + {1 \over 2} mv2

∴\therefore     hcλ1−ϕhcλ2−ϕ=12m×4v212mv2{{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m \times 4{v^2}} \over {{1 \over 2}m{v^2}}}

⇒\Rightarrow     hcλ1−ϕ{{{hc} \over {{\lambda _1}}} - \phi } = 4 (hcλ2−ϕ)\left( {{{hc} \over {{\lambda _2}}} - \phi } \right)

⇒\Rightarrow     4hcλ2{{4hc} \over {{\lambda _2}}} −- hcλ1{{hc} \over {{\lambda _1}}} = 3 ϕ\phi

⇒\Rightarrow     ϕ\phi == hc3(4λ2−1λ1){{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right)

⇒\Rightarrow     ϕ\phi == 12403(4540−1350){{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} \right)

⇒\Rightarrow     ϕ\phi = 1.8 eV

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