Question

Calculate the velocity of ejected electron from the metal surface when light of frequency 2×10^15Hz falls on the metal surface and the threshold frequency is 7×10^14hz for metal?

Answer 1

Given:

A light of frequency 2×10¹⁵ Hz falls on the metal surface and the threshold frequency is 7×10¹⁴ hz for metal.

To find:

Velocity of ejected electrons ?

Calculation:

Applying Einstein's Photo-electric Effect Equation:

$\rm \: KE = h \nu - w_{0}$

$\rm \implies\: KE = h \nu - h\nu_{0}$

$\rm \implies\: KE = h (\nu - \nu_{0})$

$\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (\nu - \nu_{0})$

$\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (20 \times {10}^{14} - 7 \times {10}^{14} )$

$\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (13 \times {10}^{14} )$

$\rm \implies\: \dfrac{1}{2}m {v}^{2} = 6.6 \times {10}^{ - 34} \times 13 \times {10}^{14}$

$\rm \implies\: {v}^{2} = \dfrac{2 \times 6.6 \times {10}^{ - 34} \times 13 \times {10}^{14} }{9.1 \times {10}^{ - 31} }$

$\rm \implies\: {v}^{2} = 18.94 \times {10}^{11}$

$\rm \implies\: v = 13.76 \times {10}^{5} \: m {s}^{ - 1}$

So, velocity of electron is 13.76 × 10⁵ m/s.

Answer 2

Answer:

$Given:</p><p></p><p>A light of frequency 2×10¹⁵ Hz falls on the metal surface and the threshold frequency is 7×10¹⁴ hz for metal.</p><p></p><p>To find:</p><p></p><p>Velocity of ejected electrons ?</p><p></p><p>Calculation:</p><p></p><p>Applying Einstein's Photo-electric Effect Equation:</p><p></p><p>\rm \: KE = h \nu - w_{0}KE=hν−w0</p><p></p><p>\rm \implies\: KE = h \nu - h\nu_{0}⟹KE=hν−hν0</p><p></p><p>\rm \implies\: KE = h (\nu - \nu_{0})⟹KE=h(ν−ν0)</p><p></p><p>\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (\nu - \nu_{0})⟹21mv2=h(ν−ν0)</p><p></p><p>\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (20 \times {10}^{14} - 7 \times {10}^{14} )⟹21mv2=h(20×1014−7×1014)</p><p></p><p>\rm \implies\: \dfrac{1}{2}m {v}^{2} = h (13 \times {10}^{14} )⟹21mv2=h(13×1014)</p><p></p><p>\rm \implies\: \dfrac{1}{2}m {v}^{2} = 6.6 \times {10}^{ - 34} \times 13 \times {10}^{14}⟹21mv2=6.6×10−34×13×1014</p><p></p><p>\rm \implies\: {v}^{2} = \dfrac{2 \times 6.6 \times {10}^{ - 34} \times 13 \times {10}^{14} }{9.1 \times {10}^{ - 31} }⟹v2=9.1×10−312×6.6×10−34×13×1014</p><p></p><p>\rm \implies\: {v}^{2} = 18.94 \times {10}^{11}⟹v2=18.94×1011</p><p></p><p>\rm \implies\: v = 13.76 \times {10}^{5} \: m {s}^{ - 1}⟹v=13.76×105ms−1</p><p></p><p>So, velocity of electron is 13.76 × 10⁵m/s.</p><p></p><p>$

solution by rohit.