Light of wavelength `lambda` from a small `0.5 mW` He-Ne laser source is used in the school laboratory to shine on perfectly absorbing surface of a spacecraft of mass `1000kg `. The tịne needed' for the spacecraft to reachi a velocity of `1 km/s` from rest is given by `alpha times 10^(beta) (s)` in scientific notation. Find the value of `(beta-r 2 alpha)`.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and apply the relevant physics concepts.

### Step 1: Determine the number of photons emitted per second by the laser.
The power of the laser is given as \( P = 0.5 \, \text{mW} = 0.5 \times 10^{-3} \, \text{W} \).

The energy of a single photon can be expressed as:
\[ E = h \nu = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the light.

The number of photons emitted per second, \( n \), can be calculated using the formula:
\[ P = n \cdot E \implies n = \frac{P}{E} = \frac{P \lambda}{hc} \]
Substituting the values, we have:
\[ n = \frac{0.5 \times 10^{-3} \cdot \lambda}{6.626 \times 10^{-34} \cdot 3 \times 10^8} \]

### Step 2: Calculate the rate of change of momentum.
The momentum of a single photon is given by:
\[ p = \frac{E}{c} = \frac{hc}{\lambda c} = \frac{h}{\lambda} \]
The rate of change of momentum (force) due to \( n \) photons is:
\[ F = n \cdot p = n \cdot \frac{h}{\lambda} \]
Substituting the expression for \( n \):
\[ F = \left(\frac{0.5 \times 10^{-3} \cdot \lambda}{6.626 \times 10^{-34} \cdot 3 \times 10^8}\right) \cdot \frac{h}{\lambda} = \frac{0.5 \times 10^{-3} \cdot 6.626 \times 10^{-34}}{6.626 \times 10^{-34} \cdot 3 \times 10^8} = \frac{0.5 \times 10^{-3}}{3 \times 10^8} \]

### Step 3: Relate force to acceleration.
Using Newton's second law, \( F = ma \), where \( m = 1000 \, \text{kg} \):
\[ a = \frac{F}{m} = \frac{0.5 \times 10^{-3}}{3 \times 10^8 \cdot 1000} \]
Calculating \( a \):
\[ a = \frac{0.5 \times 10^{-3}}{3 \times 10^{11}} = \frac{1}{6 \times 10^{11}} \, \text{m/s}^2 \]

### Step 4: Calculate the time to reach a velocity of \( 1 \, \text{km/s} \).
Using the first equation of motion \( v = u + at \) (where \( u = 0 \)):
\[ t = \frac{v}{a} = \frac{1000}{\frac{1}{6 \times 10^{11}}} = 1000 \cdot 6 \times 10^{11} = 6 \times 10^{14} \, \text{s} \]

### Step 5: Express time in scientific notation.
The time \( t \) can be expressed as:
\[ t = 6 \times 10^{14} \, \text{s} \]
Here, \( \alpha = 6 \) and \( \beta = 14 \).

### Step 6: Calculate \( \beta - 2\alpha \).
\[ \beta - 2\alpha = 14 - 2 \cdot 6 = 14 - 12 = 2 \]

### Final Answer:
The value of \( \beta - 2\alpha \) is \( 2 \).

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