Answer:
Information shadow: In the given problem, power of
the bulb is given (i.e., 25 watt). It means that the bulb emits 25 J energy per
second.
\[\therefore \] Energy released per \[\sec
=25\,J\] (i)
Wavelength of light \[(\lambda
)=0.57\mu m\]
\[=0.57\times {{10}^{-6}}m\] (ii)
Velocity of light \[(c)=3\times
{{10}^{8}}m\,{{\sec }^{-1}}\] (iii)
Problem solving strategy:
In this problem, number of quanta emitted per second is required. In order to
find it we have to calculate the number of quanta, that will provide 25 J of
energy. Let \['x'\] quanta of light are emitted per sec. Thus, energy of \['x'\]
quanta should be equal to 25 J.
\[25=x\times hv\]
\[=x\times \frac{hc}{\lambda }\] .(iv)
Working it out: From
equation (iv)
\[x=25\times \frac{\lambda
}{hc}\]
\[=25\times \frac{0.57\times
{{10}^{-6}}}{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}\]
\[=7.168\times {{10}^{19}}\]quanta per sec
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