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Step-by-Step Solution
Step 1: Calculate the total energy emitted in the given time
The power of the source is 1 mW (i.e., 1×10⁻³ J/s), and it operates for 0.1 s.
Therefore, the total energy emitted (Etotal) is:
E_\text{total} = \text{Power} \times \text{Time} = (1\times10^{-3}\,\text{J/s}) \times 0.1\,\text{s} = 1\times10^{-4}\,\text{J}.
Step 2: Express the energy of one photon
The energy of a single photon (Ephoton) with wavelength \lambda is given by Planck’s relation:
E_\text{photon} = \frac{hc}{\lambda},
where:
• h = 6.63\times10^{-34}\,\text{J·s} (Planck's constant)
• c = 3.0\times10^{8}\,\text{m/s} (speed of light)
• \lambda = 1000\,\text{nm} = 1000\times10^{-9}\,\text{m}.
Step 3: Relate total energy to the number of photons
Let n be the number of photons emitted. The total energy is then:
E_\text{total} = n \times E_\text{photon} = n \times \frac{hc}{\lambda}.
Substituting E_\text{total} = 1\times10^{-4}\,\text{J} and the values of h, c, and \lambda , we get:
10^{-4}\,\text{J} = n \times \frac{(6.63\times10^{-34}\,\text{J·s}) \times (3.0\times10^{8}\,\text{m/s})}{1000\times10^{-9}\,\text{m}}.
Step 4: Solve for the number of photons
Simplifying the expression:
10^{-4} = n \times \frac{6.63 \times10^{-34} \times3.0 \times10^{8}}{1000 \times10^{-9}}.
Calculate the denominator: 1000\times10^{-9} = 10^{-6}.
Therefore,
10^{-4} = n \times \frac{6.63 \times10^{-34} \times3.0\times10^{8}}{10^{-6}} = n \times 6.63 \times3.0 \times10^{-34+8+6}.
10^{-4} = n \times 6.63 \times3.0 \times10^{-20}.
Let's approximate 6.63\times3.0 \approx 19.9 \approx 20.0. So,
10^{-4} \approx n \times 20.0\times10^{-20} = n \times 2.0\times10^{-19}.
Hence,
n \approx \frac{10^{-4}}{2.0\times10^{-19}} = \frac{10^{-4}}{2.0}\times10^{19} = 0.5\times10^{15} = 5.0\times10^{14}.
Step 5: Express the answer in the given form
We have n \approx 5.0\times10^{14} . The question wants the number in the form x\times10^{13} .
So we can rewrite 5.0\times10^{14} = 50.0\times10^{13}. Thus, x = 50. (nearest integer)
Final Answer
The value of x is 50.
