Given: - Frequency of light: \( \nu = 6 \times 10^{14} \, \text{Hz} \) - Power emitted by the source: \( P = 2 \times 10^{-3} \, \text{W} \) - Planck's constant: \( h = 6.63 \times 10^{-34} \, \text{Js} \)
The energy \( E \) of a photon is given by:
\[ E = h\nu \]
Substituting the given values:
\[ E = 6.63 \times 10^{-34} \times 6 \times 10^{14} \, \text{J} \] \[ E = 3.978 \times 10^{-19} \, \text{J} \]
Rounding off:
\[ E \approx 4 \times 10^{-19} \, \text{J} \]
The number of photons emitted per second (\( n \)) is given by:
\[ n = \frac{P}{E} \]
Substituting the given values:
\[ n = \frac{2 \times 10^{-3}}{4 \times 10^{-19}} \] \[ n = \frac{2}{4} \times 10^{16} \] \[ n = 0.5 \times 10^{16} \] \[ n = 5 \times 10^{15} \]
Conclusion: The number of photons emitted per second by the source is \( 5 \times 10^{15} \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: