Science, asked by souravo2071, 1 year ago

A photon of wavelength 250 nm ejects an electron from a metal. The ejected electron has a de Broglie wavelength of 0.85 nm.

Answers

Answered by ankitsharma26
10
(a) p = h / λ … mv = h / λ … (9.11 x 10–31)v = 6.63 x 10–34 / 0.85 x 10–9 … v = 8.56 x 105 m/s

K = ½ mv2 = ½ (9.11 x 10–31)(8.56 x 105)2 = 3.34 x 10–19 J

(b) K = Ein – φ … K = hc / λ – φ … 3.34 x 10–19 = 1.99 x 10–25 / 250 x 10–9 – φ … φ = 4.62 x 10–19 J = 2.89 eV

(c) From E = hc / λ, a bigger λ means a smaller energy. The X emission of 400 nm is a smaller energy than the 250 nm photon. Since the photon was “created” from this transition, it must be an emission so we should go down energy levels with an energy difference larger than the X level difference. This would be transition d.
Answered by Sweetbuddy
0
HEY BUDDY HERE IS UR ANSWER !!

(a) p = h / λ
mv = h / λ (9.11 x 10–31)
v = 6.63 x 10–34 / 0.85 x 10–9
v = 8.56 x 105 m/s

K = ½ mv2 = ½ (9.11 x 10–31)(8.56 x 105)2 = 3.34 x 10–19 J

(b) K = Ein – φ … K = hc / λ – φ ..3.34 x 10–19 = 1.99 x 10–25 / 250 x 10–9 – φ … φ = 4.62 x 10–19 J = 2.89 eV

(c) From E = hc / λ, a bigger λ means a smaller energy. The X emission of 400 nm is a smaller energy than the 250 nm photon. Since the photon was “created” from this transition, it must be an emission so we should go down energy levels with an energy difference larger than the X level difference. This would be transition d.


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