When light of wavelength 248 nm falls on a metal of threshold energy 3.0eV, the de-Broglie wavelength of emitted electrons is__________ Angstrom.
When light of wavelength 248 nm falls on a metal of threshold energy 3.0eV, the de-Broglie wavelength of emitted electrons is__________ Angstrom.
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1 Answer
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K.E = φ - φ?
φ? = 3 eV = 3 × 1.6 × 10? ¹? J = 4.8 × 10? ¹? J
φ = hc/λ = (6.63 × 10? ³? × 3 × 10? ) / (248 × 10? ) J = 8 × 10? ¹? J
K.E = 8 × 10? ¹? - 4.8 × 10? ¹? = 3.2 × 10? ¹? J
Now using, λ = h / √ (2 K.E m)
λ = (6.63 × 10? ³? ) / √ (2 × 3.2 × 10? ¹? × 9.1 × 10? ³¹) m
λ = (6.63 × 10? ³? ) / (7.63 × 10? ²? ) m = 0.87 × 10? m = 8.7 Å
So, the nearest integer is 9.
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