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Chemistry Question 71 – JEE-MAIN 2026

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The surface of sodium metal is irradiated with radiation of wavelength x nm. The kinetic energy of ejected electrons is 2.8×10−20 J. The work function of sodium is 2.3 eV. The value of x is _______ ×102 nm. (Nearest integer) (Given: h=6.6×10−34 J s; 1 eV =1.6×10−19 J; c=3.0×108 m s−1)

The energy of the incident photon is used to overcome the work function of the metal and provide kinetic energy to the ejected electrons.

Step 1: Convert Work Function and Calculate Total Photon Energy✦ Active

The work function ϕ is given in electron volts (eV), so convert it to Joules (J) using the conversion factor 1 eV=1.6×10−19 J.

ϕ=2.3 eV×1.6×10−19 J/eV=3.68×10−19 J

According to Einstein's photoelectric equation, the energy of the incident photon E is the sum of the work function and the kinetic energy of the ejected electron.

E=ϕ+K.E.=3.68×10−19 J+2.8×10−20 J=3.96×10−19 J
Step 2: Calculate Wavelength of Incident Radiation○ Expand

The energy of a photon is also related to its wavelength λ by the formula E=hcλ. We can rearrange this to solve for λ.

λ=hcE=(6.6×10−34 J s)×(3.0×108 m s−1)3.96×10−19 J
λ=19.8×10−263.96×10−19 m=5.0×10−7 m
Step 3: Convert Wavelength to Nanometers and Find x○ Expand

Convert the wavelength from meters to nanometers, knowing that 1 m=109 nm.

λ=5.0×10−7 m×109 nm1 m=500 nm

The question asks for the value of x such that the wavelength is x×102 nm.

500 nm=5×102 nm

Therefore, x=5.

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