The photosensitive metal is first irradiated with the radiation of wavelength 400 mm and then with radiation of wavelength 800 nm. The change in the maximum kinetic energy of the photoelectron assuming that both wavelengths can cause photoelectric emission, is

0.55 eV

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1.55 eV

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20.eV

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1.0 eV

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Solution

The correct option is B
1.55 eV

$E_{k_{1}} - E_{k_{2}} = E_{1} - E_{2} S i n c e λ_{2} = 2 λ_{1}, the photon energies are related as E_{2} = \frac{E_{1}}{2} \Rightarrow Δ E_{k} = E_{1} = \frac{E_{1}}{2} = \frac{E_{1}}{2} E (i n e V) ≅ \frac{12420}{λ (˙ A)} = 3.1 e V E_{2} = \frac{E_{1}}{2} = 1.55 e V \Rightarrow E_{1} - E_{2} = Δ E ≅ 1.55 e V$

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The photosensitive metal is first irradiated with the radiation of wavelength 400 mm and then with radiation of wavelength 800 nm. The change in the maximum kinetic energy of the photoelectron assuming that both wavelengths can cause photoelectric emission, is

The correct option is B 1.55 eV Ek1−Ek2=E1−E2Since λ2=2λ1, the photon energies are related as E2=E12⇒ΔEk=E1=E12=E12E(in eV)≅12420λ(˙A)=3.1 eV E2=E12=1.55eV⇒E1−E2=ΔE≅1.55eV