Question

light of wavelength 400 nm strikes a certain metal which has a photoelectric work function of 2.13ev find out maximum kinetic energy of the photoelectrons​

Answer 1

Given :

• Wavelength = 400 nm
• Work function = 2.13eV

To find :

Maximum Kinetic energy (KE) of photoelectron

Formula used :

$\sf \bullet \: \: KE_{max} \: = \dfrac{hc}{ \lambda} - W_0$

Here,

• h = Planck's constant

• c = speed of light

$\sf \bullet\: KE_{max}\: =\: kinetic\: energy \\\\ \sf \bullet \lambda \:=\: wavelength \\ \\ \sf \bullet W_{0} \: = \: work\: function$

Solution :

Putting given value in above formula, we get;

$\sf \implies \: KE_{max} \: = (\dfrac{ \: \: 6.626 \times {10}^{ - 34} \times \: 3 \: \times \: {10}^{8} \: }{400 \times {10}^{ - 9} } \: ) - \: (2.13 \times 1.6 \times {10}^{ - 19} )\\ \\ \sf \implies \: KE_{max} \: = (\dfrac{19.878}{4} \times {10}^{ - 19} ) \: - \: (3.408 \times {10}^{ - 19} ) \\ \\ \sf \implies \: KE_{max} \: = (4.9695 \times {10}^{ - 19} ) - \: (3.408 \times {10}^{ - 19} ) \\ \\ \sf \implies \: KE_{max} \: = (4.9695 \: - \: 3.408) \: \times {10}^{ - 19} \\ \\ \sf \implies \: KE_{max} \: = 1.5651 \: \times \: {10}^{ - 19} \: J \\ \\ \sf \implies \: KE_{max} \: \approx \: 1.57 \times {10 }^{ - 19} \: J$

ANSWER :

Maximum Kinetic energy of photoelectron = $\sf \bold{1.57\: \times 10^{-19}}$

Answer 2

Answer:

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Explanation:

Putting given value in above formula, we get;

\begin{gathered}\sf \implies \: KE_{max} \: = (\dfrac{ \: \: 6.626 \times {10}^{ - 34} \times \: 3 \: \times \: {10}^{8} \: }{400 \times {10}^{ - 9} } \: ) - \: (2.13 \times 1.6 \times {10}^{ - 19} )\\ \\ \sf \implies \: KE_{max} \: = (\dfrac{19.878}{4} \times {10}^{ - 19} ) \: - \: (3.408 \times {10}^{ - 19} ) \\ \\ \sf \implies \: KE_{max} \: = (4.9695 \times {10}^{ - 19} ) - \: (3.408 \times {10}^{ - 19} ) \\ \\ \sf \implies \: KE_{max} \: = (4.9695 \: - \: 3.408) \: \times {10}^{ - 19} \\ \\ \sf \implies \: KE_{max} \: = 1.5651 \: \times \: {10}^{ - 19} \: J \\ \\ \sf \implies \: KE_{max} \: \approx \: 1.57 \times {10 }^{ - 19} \: J\end{gathered}

⟹KE

max

=(

400×10

−9

6.626×10

−34

×3×10

8

)−(2.13×1.6×10

−19

)

⟹KE

max

=(

4

19.878

×10

−19

)−(3.408×10

−19

)

⟹KE

max

=(4.9695×10

−19

)−(3.408×10

−19

)

⟹KE

max

=(4.9695−3.408)×10

−19

⟹KE

max

=1.5651×10

−19

J

⟹KE

max

≈1.57×10

−19

J

ANSWER :

Maximum Kinetic energy of photoelectron = \sf \bold{1.57\: \times 10^{-19}}1.57×10

−19