NEET Questions Solved

NEET - 2016

When a metallic surface is illuminated with radition of wavelength λ, the stopping potential is V, if the same surface is liiuminated with radiation of wavelength 2λ, the stopping potential is V4 The threshold wavelength for matallic surface is 

(a) 5λ                (b)52λ

(c) 3λ                (d) 4λ

(c) In Ist case, when a matallic surface is illuminated with radiation of wavelength λ, the stopping potential is V.

So, photoelectric equation can be written as

              eV=hcλ=hcλο                        ...(i)

In IInd Case, when the same surface is illuminayed with radition of wavelength 2λ, the sopping potential is V4 so, photoelectric equation can be written as

                        eV4=hc2λ-hcλ0

                eV=4hc2π-4hcλ0                    ...(ii)

From eqs, (i) and (ii) we get

              hcλ-hcλ0=4hc2λ-4hcλο              1λ-1λο=  2λ-  4λο                 λ0=3λ    

Difficulty Level:

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