JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is $1 0^{16} s^{- 1}$ . The wave number of photon 'C' is $1 0^{4} cm^{- 1}$ . The correct order of energy of these photons is :

A
C $>$ B $>$ A
B
B $>$ A $>$ C
C
A $>$ B $>$ C
D
A $>$ C $>$ B

Solution

Convert each to a common comparable quantity (wavelength):

$λ_{A} = 400$ nm.

$λ_{B} = \frac{c}{ν} = \frac{3 \times 1 0 ^{8}}{1 0 ^{16}} = 3 \times 1 0^{- 8}$ m $= 30$ nm.

$λ_{C} = \frac{1}{ν ˉ} = \frac{1}{1 0 ^{4} cm ^{- 1}} = 1 0^{- 4}$ cm $= 1 0^{- 6}$ m $= 1000$ nm.

So $λ_{B} < λ_{A} < λ_{C}$ . Since $E \propto 1/ λ$ : $E_{B} > E_{A} > E_{C}$ .

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