    Question

# In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectric energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light at the corresponding stopping potential (V0) are given below: λ(μm) V0 (Volt) 0.3 2.0 0.4 1.0 0.5 0.4 Given, c=3×108 ms−1 and e=1.6×10−19 C, Planck's constant (in units of J−s) from such an experiment is

A
6.0×1034
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B
6.4×1034
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C
6.6×1034
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D
6.8×1034
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Solution

## The correct option is B 6.4×10−34Einstein's Photoelectric equation hcλ−ϕ=eV0 From the given data; for first wavelenghthc0.3×10−6−ϕ=2e .....(1) for secound wavelenghthc0.4×10−6−ϕ=e .....(2) Subtracting (1) to (2), we get, ⇒hc(10.3−10.4)106=e where, c=3×108 ms−1 and e=1.6×10−19 C ⇒h×(3×108)(10.3−10.4)106=1.6×10−19 h=6.4×10−34 Js  Suggest Corrections  2      Explore more