The surface of certain metal is first illuminated with light of wavelength λ _ 1 = 350nm and then, by a light of wavelength λ _ 2 = 540nm . It is found that the maximum speed of the photoelectrons in the two cases

Correct answer: B. 1.8 λ _ 1 = 350nm, λ _ 2 = 540nm, V _ 1 V _ 2 = 2 ⇒ V _ 1 = 2Vand V _ 2 = V hc λ _ 1 = ϕ + (1)/(2)m (2V) ^ 2 hc λ _ 2 = ϕ + (1)/(2)m V ^ 2 hc λ _ 1 - 4 hc λ _ 2 = ϕ - 4ϕ hc((4)/(540) - (1)/(340)) = 3ϕ ϕ = 1.8eV

JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2019 · 9 January · Morning Shift · Physics · Dual Nature of Matter

The surface of certain metal is first illuminated with light of...

The surface of certain metal is first illuminated with light of wavelength

{\lambda}_{1} = 350nm and then, by a light of wavelength

{\lambda}_{2} = 540nm .

It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of

. The work function of the metal (in

) is close to<br /><br>(Energy of photon

= \dfrac{1240}{\lambda

\left(innm\right)}eV

)

Options

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

Correct answer

B. $1.8$

Step-by-step solution

{\lambda}_{1} = 350nm, {\lambda}_{2} = 540nm, \dfrac{{V}_{1}}{{V}_{2}} = 2 ⇒ {\text{V}}_{1} = 2\text{V}and{\text{V}}_{2} = V

\dfrac{hc}{{\lambda}_{1}} = ϕ + \dfrac{1}{2}m{\left(2V\right)}^{2}

\dfrac{hc}{{\lambda}_{2}} = ϕ + \dfrac{1}{2}m{V}^{2}

\dfrac{hc}{{\lambda}_{1}} - 4\dfrac{hc}{{\lambda}_{2}} = ϕ - 4ϕ

hc\left(\dfrac{4}{540} - \dfrac{1}{340}\right) = 3ϕ

ϕ = 1.8eV

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