Calculate the number of photons emitted by a `100 W` yellow lamp in `1.0 s`. Take the wavelength of yellow light as `560 nm` and assume `100` percent efficiency.
A. `6.8xx10^(20)`
B. `4xx10^(12)`
C. `4xx10^(20)`
D. `2.8xx10^(20)`
Correct Answer - 4
The number pf photon is `N=E/(hv)=(PDeltat)/(h(c//lambda))=(lambdaPDeltat)/(hc)`
Substitution of the data gives
`N=((5.60xx10^(-7)m)xx(100Js^(-1))xx(1.0s))/((6.626xx10^(-36)s)xx(3xx10^(8)ms^(-1)))=2.8xx10^(20)`
Correct Answer - A::C::D
Room temperature `rArr n = 1`
Upon absorption excitations takes place to many higher states which upon de-excitation emit all U.V., infrared and visible light.
Correct Answer - (i)`125/33 muF` or `2.4 H` , (ii)`720 Omega` , (iii)it will be more economical to use inductance or capacitance in series with the lamp to run it...
Correct Answer - `1xx10^(16) Hz`
For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)`
`:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)`
For frequency `4xx10^(16)` Hz,
`h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)`
Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....