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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2026 · 23 January · Evening Shift · Chemistry · Structure of Atom

The work functions of two metals (M_ A . and .M_ B ) are in the 1:...

The work functions of two metals (MAandMB)areinthe\left(M_{A}\right. and \left.M_{B}\right) are in the1: 2ratio.Whenthesemetalsareexposedtophotonsofenergy6eV,thekineticenergyofliberatedelectronsofratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons ofM_{A}: M_{B}isintheratioofis in the ratio of2.642: 1.Theworkfunctions(ineV). The work functions (in eV) of M_{A} and M_{B}$ are respectively.

Options

  1. A1.4,2.81.4,2.8
  2. B2.3,4.62.3,4.6
  3. C1.5,3.01.5,3.0
  4. D3.1,6.23.1,6.2

Correct answer

B. 2.3,4.62.3,4.6

Step-by-step solution

The work functions satisfy WA:WB=1:2W_A : W_B = 1:2.
Let WA=WW_A = W and WB=2WW_B = 2W.
Using the photoelectric equation
KE=hν−WKE = h\nu - W, we have KEA=6−WKE_A = 6 - W and KEB=6−2WKE_B = 6 - 2W.
Given the kinetic energy ratio KEAKEB=2.642\frac{KE_A}{KE_B} = 2.642, we get 6−W6−2W=2.642\frac{6-W}{6-2W} = 2.642.
Solving: 6−W=2.642(6−2W)6 - W = 2.642(6 - 2W) gives 6−W=15.852−5.284W6 - W = 15.852 - 5.284W
So 4.284W=9.8524.284W = 9.852 and W=2.3W = 2.3 eV.
Thus WA=2.3W_A = 2.3 eV and WB=4.6W_B = 4.6 eV.
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