Physics, asked by ebinfibin, 9 months ago

the threshold frequency for a metal is 6.2×10.^3 8-1.calculate KE emitted of an electron when the radiation of frequency v=8.7×10^14 S-1 strikes the metal.​

Answers

Answered by Rajshuklakld
6

Threshold frequency:-This is the frequency of energy upto which electron doesn't able to overcome the nuclear energy....

Threshold energy:-Energy upto which electron are not able to break the nuclear force ,is the threshold energy

Solution:-

threshold \: energy = h \times vo \\ where \: h \: is \: the \: planks \: constant  \: and\\vo \: is \: the \: threshold \: frequency  \:  \\ threshold \: energy = 6.626 \times  {10}^{ - 34} \times 6.2 \times  {10}^{ - 38}   \\  = 9.6 \times  {10}^{ - 72} \\ frequency \: of \: radiation = 8.7 \times  {10}^{14} \times 6.626 \times  {10}^{ - 34}  \\  =  57.42 \times  {10}^{ - 20} \\ energy \: transferred \: by \: radiation = 6.6 \times  {10}^{ - 34}  \times 57.42 \times  {10}^{ - 20} \\ 3.8 \times  {10}^{ - 12}  \\ kinetic \: energy = total \: energy \: transferred -  \\ threshold \: energy \\ kinetic \: energy = 3.8 \times  {10}^{ - 12}  - 9.6 \times  {10}^{ - 72} \\ since \: 9.8 \times  {10}^{ - 72} is very \: very \: small \:  \\ so \: we \: will \: neglect \: it \\ hence \: 3.8 \times  {10}^{ - 12}  \: is \: the \: required \: kinetic \: energy

Answered by ss6255944
0

Answer:

Threshold frequency:-This is the frequency of energy upto which electron doesn't able to overcome the nuclear energy....

Threshold energy:-Energy upto which electron are not able to break the nuclear force ,is the threshold energy

Solution:-

\begin{gathered}threshold \: energy = h \times vo \\ where \: h \: is \: the \: planks \: constant \: and\\vo \: is \: the \: threshold \: frequency \: \\ threshold \: energy = 6.626 \times {10}^{ - 34} \times 6.2 \times {10}^{ - 38} \\ = 9.6 \times {10}^{ - 72} \\ frequency \: of \: radiation = 8.7 \times {10}^{14} \times 6.626 \times {10}^{ - 34} \\ = 57.42 \times {10}^{ - 20} \\ energy \: transferred \: by \: radiation = 6.6 \times {10}^{ - 34} \times 57.42 \times {10}^{ - 20} \\ 3.8 \times {10}^{ - 12} \\ kinetic \: energy = total \: energy \: transferred - \\ threshold \: energy \\ kinetic \: energy = 3.8 \times {10}^{ - 12} - 9.6 \times {10}^{ - 72} \\ since \: 9.8 \times {10}^{ - 72} is very \: very \: small \: \\ so \: we \: will \: neglect \: it \\ hence \: 3.8 \times {10}^{ - 12} \: is \: the \: required \: kinetic \: energy\end{gathered}

thresholdenergy=h×vo

wherehistheplanksconstantand

voisthethresholdfrequency

thresholdenergy=6.626×10

−34

×6.2×10

−38

=9.6×10

−72

frequencyofradiation=8.7×10

14

×6.626×10

−34

=57.42×10

−20

energytransferredbyradiation=6.6×10

−34

×57.42×10

−20

3.8×10

−12

kineticenergy=totalenergytransferred−

thresholdenergy

kineticenergy=3.8×10

−12

−9.6×10

−72

since9.8×10

−72

isveryverysmall

sowewillneglectit

hence3.8×10

−12

istherequiredkineticenergy

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