When a light of wavelength $400 n m$ falls on the metal of work function $2.5 e V .$ What will be the maximum magnitude of linear momentum of emitted photoelectron?

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Solution

Energy of incident photon is $E = \frac{1240}{λ (n m)} e V = \frac{1240}{400} = 3.1 e V$

And work function is $W = 2.5 e V$

So maximum energy of ejected electron will be, $K E = E - W = 3.1 e V - 2.5 e V = 0.6 e V = 0.6 \times 1.6 \times 10^{- 19} J o u l e$

Maximum possible linear momentum will be $p = \sqrt[2]{2 m K E} = \sqrt[2]{2 \times 9.1 \times 10^{- 31} K g \times 0.96 \times 10^{- 19} J o u l e} = 4.12 \times 10^{- 25} K g m / s$

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When a light of wavelength 400nm falls on the metal of work function 2.5eV. What will be the maximum magnitude of linear momentum of emitted photoelectron?

Energy of incident photon is E=1240λ(nm)eV=1240400=3.1eVAnd work function is W=2.5eVSo maximum energy of ejected electron will be, KE=E−W=3.1eV−2.5eV=0.6eV=0.6×1.6×10−19JouleMaximum possible linear momentum will be p=2√2mKE=2√2×9.1×10−31Kg×0.96×10−19Joule=4.12×10−25Kgm/s

When a light of wavelength $400 n m$ falls on the metal of work function $2.5 e V .$ What will be the maximum magnitude of linear momentum of emitted photoelectron?