Answer :

Given

Wavelength of incident light λ = 412.5nm = 412.5× 10^{-9}m

Energy of photon of light of a given frequency is given as

E = hν

Where h = Planck’s constant = 6.63× 10^{-34}Js

ν = frequency of light

We know that

Where c = speed of light = 3× 10^{8}ms

We know that 1eV = 1.60× 10^{-19}J.

Substituting the value of λ , h and c in above equation we get,

Photoelectric emission will only take place if the energy of the photon of the incident radiation is more than the work function of the given metal.

From the table given in the question, we can see that work function of Na (1.92eV) and K (2.15eV) is less than the energy of photon of incident radiation. Therefore, metals Na and K will show photoelectric emission.

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