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
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:-
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