InterviewSolution
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InterviewSolution
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Neon gas is generally used in the sign boards. If it emits strongly at `616 nm`, calculate a. The frequency of emission, b. The distance traveled by this radiation in `30 s` c. The energy of quantum and d. The number of quanta present if it produces `2 J` of energy. |
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Answer» Wavelength of radiation emitted = 616 nm` = 616 × 10^(-9) m` (Given) (a) Frequency of emission (v) `v = (c)./(lambda)` c = velocity of radiation `lambda =` wavelength of radiation Substituting the values in the given expression of (v) : `v = (3.0 xx 10^(8) m//s)/(616 xx 10^(-9) m)` `= 4.87 xx 10^(8) xx 10^(9) xx 10^(-3) s^(-1)` `v = 4.87 xx 10^(14) s^(-1)` Frequency of emission `(v) = 4.87 xx 10^(14) s^(-1)` (b) Velocity of radiation, `(c). = 3.0 × 10^(8) ms^(1)` Distance travelled by this radiation in 30 s `= (3.0 xx 10^(8)ms^(-1))(30s)` `= 9.0 xx 10^(9) m` (c). Energy of quantum `(E) = hν` `(6.626 xx 10^(-34)Js)(4.87 xx 10^(14)s^(-1))` Energy of quantum `(E) = 32.87 xx 10^(-20) J` (d) Energy of one photon (quantum) `= 32.27 xx 10^(-20) J` Therefore, `32.27 xx 10^(-20) J` of energy is present in 1 quantum. Number of quanta in 2 J of energy `= (2J)/(32.27 xx 10^(-20) J)` `= 6.19 xx 10^(18)` `= 6.2 xx 10^(18)` |
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