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When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ______ m. (h=6.63×10⁻³⁴ J s, e=1.6×10⁻¹⁹ C, c=3×10⁸ m/s)

Asked in JEE Main 24th Jan 2nd Shift 2026 · Stopping potential from wavelength

• (1) 3.1 imes10⁻⁷
• (2) 2.9 imes10⁻⁸
• (3) 2.2 imes10⁻⁸
• (4) 2.5 imes10⁻⁷

Step-by-step solution

Why the other options are wrong

• (1) comes from rounding the potential difference 2.5 V down to 2 V, giving λ=hc/4e
• (2) comes from an exponent slip in hc, which drops the answer a whole decade; 29 nm is soft X-ray territory, not a photocell source
• (3) comes from the same lost decade in hc; any answer near 10⁻⁸ m should be checked against the 10⁻⁷ m scale of visible and UV light

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