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.
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
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