Question

calculate the threshold frequency of the metal from which the photoelectrons are emitted with zero velocity when exposed to radiation of wavelength 6800armstongs

Answer 1

Given λ = 6800 amgstrom or 6800x 10-10 m = 6.8 x 10-7m

C = 3 x 108 m/s

Also λ = c/v or v = c/ λ Therefore putting the values in the equation, we get

=3 x 108 / 6.8 x 10-7m = 4.41 x 1014 s-1

Hence, work function (W0) of the metal = hν0

= (6.626 × 10–34 Js) (4.41 × 1014 s–1)

= 2.922 × 10–19 J

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Answer 2

The threshold frequency of the metal from which the photoelectrons are emitted with zero velocity when exposed to radiation of wavelength 6800armstongs is

c = νλ ∴ νo = c/λo = 3.0 × 108 ms-1/6800×10-10 m = 4.14 × 1014 s-1

Work function (wo) = hνo = 6.626×10-34 Js×4.14×1014 s-1 = 2.92×10-19 J