Light of wavelength 500 nm is incident on a metal with a work 2-28 eV- The de Borglie wavelength of the emitted electron is- 2-8 times 10-9-mleq 2-8 times 10-12-mgeq 2-8 times 10-9-m- 2-8 times 10-10-m

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Toppr Admin · Accepted Answer

Energy of the photon- E-124005000-2-48eVWork function- -x3D5-2-28eVKEmax-E-x2212-x3D5-2-48-x2212-2-28-0-2eVFor electron- -x3BB-min-h-x221A-2m-KE-max-28A0Hence- -x3BB-gt-28A0-2-8-xD7-10-x2212-9m

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:
$< 2.8 \times 10^{- 9} m$
$\leq 2.8 \times 10^{- 12} m$
$\geq 2.8 \times 10^{- 9} m$
$< 2.8 \times 10^{- 10} m$

Energy of the photon: $E = \frac{12400}{5000} = 2.48 e V$
Work function: $ϕ = 2.28 e V$
$K E_{m a x} = E - ϕ = 2.48 - 2.28 = 0.2 e V$
For electron, $λ_{m i n} = \frac{h}{\sqrt{2 m (K E)_{m a x}}} = 28 A^{0}$
Hence, $λ > 28 A^{0} = 2.8 \times 10^{- 9} m$

Light of wavelength 500 nm is incident on a metal with a work function 2.28 eV. The de Borglie wavelength of the emitted electron is:<2.8×10−9m≤2.8×10−12m≥2.8×10−9m<2.8×10−10m

Energy of the photon: E=124005000=2.48eVWork function: ϕ=2.28eVKEmax=E−ϕ=2.48−2.28=0.2eVFor electron, λmin=h√2m(KE)max=28A0Hence, λ>28A0=2.8×10−9m

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:
$< 2.8 \times 10^{- 9} m$
$\leq 2.8 \times 10^{- 12} m$
$\geq 2.8 \times 10^{- 9} m$
$< 2.8 \times 10^{- 10} m$

Energy of the photon: $E = \frac{12400}{5000} = 2.48 e V$
Work function: $ϕ = 2.28 e V$
$K E_{m a x} = E - ϕ = 2.48 - 2.28 = 0.2 e V$
For electron, $λ_{m i n} = \frac{h}{\sqrt{2 m (K E)_{m a x}}} = 28 A^{0}$
Hence, $λ > 28 A^{0} = 2.8 \times 10^{- 9} m$