Energy required to remove an electron from aluminium surface is 4-2 eV- If light of wavelength 2000 A falls on the surface- the velocity of the fastest electron ejected from the surface will beA- 7-4 - 105 m - sec- 8-4 - 105 m - sec- 6-4 - 105 m - secD- 8-4 - 105 m - sec

The correct option is B 8.4 × 10^5 m/sec By using E=W_0 + 1/2mv^2_max where E=12375/2000=6.18eV ⇒ 6.18 eV= 4.2eV + 1/2mv^2_max⇒ 1.98 eV=1/2mv^2_max ⇒ 1.98 × 1. ...

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Standard XII

Physics

The Photon Model

Energy requir...

Question

Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength $2000 \circ A$ falls on the surface, the velocity of the fastest electron ejected from the surface will be

$8.4 \times 10^{5} m / s e c$

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$7.4 \times 10^{5} m / s e c$

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$6.4 \times 10^{5} m / s e c$

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$8.4 \times 10^{5} m / s e c$

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Solution

The correct option is A
$8.4 \times 10^{5} m / s e c$

$B y u s i n g E = W_{0} + \frac{1}{2} m v_{m a x}^{2} w h e r e E = \frac{12375}{2000} = 6.18 e V$
$\Rightarrow 6.18 e V = 4.2 e V + \frac{1}{2} m v_{m a x}^{2} \Rightarrow 1.98 e V = \frac{1}{2} m v_{m a x}^{2}$
$\Rightarrow 1.98 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v_{m a x}^{2}$
$\Rightarrow v_{m a x} = 8.4 \times 10^{5} m / s$

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Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000∘A falls on the surface, the velocity of the fastest electron ejected from the surface will be

The correct option is A 8.4×105 m/sec By using E=W0+12mv2max where E=123752000=6.18eV ⇒6.18eV=4.2eV+12mv2max⇒1.98eV=12mv2max ⇒1.98×1.6×10−19=12×9.1×10−31×v2max ⇒vmax=8.4×105 m/s

Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength $2000 \circ A$ falls on the surface, the velocity of the fastest electron ejected from the surface will be