A photon of energy ‘$$2$$’ ev is incident on a metal, The minimum $$de-Broglie$$ wave length of the emitted electrons is λ. When the wave length of the incident radiation is reduced by $$60$$% the minimum $$de-Broglie$$ wave length of the emitted electrons is λ$$2$$ The work function of the metal is

Question

A photon of energy ‘$$2$$’ ev is incident on a metal, The minimum $$de-Broglie$$ wave length of the emitted electrons is λ. When the wave length of the incident radiation is reduced by $$60$$% the minimum $$de-Broglie$$ wave length of the emitted electrons is λ$$2$$ The work function of the metal is

Infinity Learn · Accepted Answer

if de Broglie wavelength becomes halved momentum becomes doubled and kinetic energy becomes $$4$$ times  and if incident wavelength decreases by $$60$$%  it means it becomes $$40$$%  λ becomes $$410$$λ  frequency becomes $$104$$ν Eienstein equation hν $$=w+k----1$$            second time h $$104$$ν $$=$$ $$w+4k----2$$ solving $$w=0.5$$ hν  here hν$$=$$ $$2ev$$

A photon of energy ‘2’ ev is incident on a metal, The minimum de-Broglie wave length of the emitted electrons is $λ$ . When the wave length of the incident radiation is reduced by 60% the minimum de-Broglie wave length of the emitted electrons is $\frac{λ}{2}$ The work function of the metal is

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A photon of energy ‘2’ ev is incident on a metal, The minimum de-Broglie wave length of the emitted electrons is λ. When the wave length of the incident radiation is reduced by 60% the minimum de-Broglie wave length of the emitted electrons is λ2 The work function of the metal is

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