Question

A 250W bulb emit light of wavelength 19.6nm. Find the no of electron emitted per second

Hint:

To simplify calculations, you can use the value $hc \approx 1240 \text{ eV}\cdot\text{nm}$. Convert Power to eV/s first ($250 \text{ J/s} = 250 / (1.6 \times 10^{-19}) \text{ eV/s}$). Then $n = \text{Total Energy in eV} / \text{Energy of one photon in eV}$.

Answer 1

Step 1: Understanding the Concept:
Power ($P$) is the total energy emitted per unit time. This energy is carried by discrete packets called photons. By calculating the energy of a single photon, we can determine how many such photons are required to make up the total power output.

Step 2: Key Formula or Approach:
Energy of a single photon: $E = \frac{hc}{\lambda}$
Total power: $P = n \times E$, where $n$ is the number of photons emitted per second.
Rearranging for $n$:
\[ n = \frac{P}{E} = \frac{P \lambda}{hc} \]

Step 3: Detailed Explanation:
Given values:
Power, $P = 250$ W = $250$ J/s
Wavelength, $\lambda = 19.6$ nm $= 19.6 \times 10^{-9}$ m
Planck's constant, $h \approx 6.626 \times 10^{-34}$ J$\cdot$s
Speed of light, $c \approx 3 \times 10^8$ m/s
Substitute these values into the formula:
\[ n = \frac{250 \times 19.6 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} \]
Using the approximation $hc \approx 19.8 \times 10^{-26}$ J$\cdot$m or more precisely $19.878 \times 10^{-26}$ J$\cdot$m:
\[ n = \frac{4900 \times 10^{-9}}{19.878 \times 10^{-26}} \]
\[ n = \left(\frac{4900}{19.878}\right) \times 10^{17} \]
\[ n \approx 246.5 \times 10^{17} \]
\[ n \approx 2.465 \times 10^{19} \]

Step 4: Final Answer:
The number of photons emitted per second is approximately $2.47 \times 10^{19}$.