The escape speed of an electron launched from the surface of a 1cm diameter glass sphere that has been charged to `10 nC` is ……. `xx 10^(7) m//sec`. (Round off in the nearest integer)

Explanation

To find the escape speed of an electron launched from the surface of a 1 cm diameter glass sphere charged to 10 nC, we can use the principle of energy conservation, where the kinetic energy of the electron equals the electric potential energy due to the charged sphere.

### Step-by-Step Solution:

1. **Identify the relevant formulas**:
- The kinetic energy (KE) of the electron is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The electric potential energy (PE) due to a point charge is given by:
\[ PE = \frac{k \cdot Q \cdot e}{r} \]
where:
- \( k \) is Coulomb's constant \( (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
- \( Q \) is the charge of the sphere \( (10 \, \text{nC} = 10 \times 10^{-9} \, \text{C}) \)
- \( e \) is the charge of an electron \( (1.6 \times 10^{-19} \, \text{C}) \)
- \( r \) is the radius of the sphere \( (0.5 \, \text{cm} = 0.5 \times 10^{-2} \, \text{m}) \)

2. **Set the kinetic energy equal to the potential energy**:
\[ \frac{1}{2} mv^2 = \frac{k \cdot Q \cdot e}{r} \]

3. **Rearranging for velocity \( v \)**:
\[ v = \sqrt{\frac{2k \cdot Q \cdot e}{m \cdot r}} \]

4. **Substituting the values**:
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Substitute \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q = 10 \times 10^{-9} \, \text{C} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), and \( r = 0.5 \times 10^{-2} \, \text{m} \):
\[ v = \sqrt{\frac{2 \cdot (9 \times 10^9) \cdot (10 \times 10^{-9}) \cdot (1.6 \times 10^{-19})}{(9.1 \times 10^{-31}) \cdot (0.5 \times 10^{-2})}} \]

5. **Calculating the numerator**:
\[ 2 \cdot (9 \times 10^9) \cdot (10 \times 10^{-9}) \cdot (1.6 \times 10^{-19}) = 2.88 \times 10^{-8} \, \text{J} \]

6. **Calculating the denominator**:
\[ (9.1 \times 10^{-31}) \cdot (0.5 \times 10^{-2}) = 4.55 \times 10^{-33} \, \text{kg m} \]

7. **Calculating the escape velocity**:
\[ v = \sqrt{\frac{2.88 \times 10^{-8}}{4.55 \times 10^{-33}}} \approx \sqrt{6.33 \times 10^{24}} \approx 7.95 \times 10^7 \, \text{m/s} \]

8. **Rounding off**:
Rounding \( 7.95 \times 10^7 \) gives approximately \( 8 \times 10^7 \, \text{m/s} \).

### Final Answer:
The escape speed of the electron is approximately \( 8 \times 10^7 \, \text{m/s} \).