In an experiment with high energy beam, hydrogen ions each of `1.67 xx 10^(-27)` kg strike a stationary and thermally insulated target with a velocity of `2xx 10^7 ms^(-1)` at the rate of `10^15` ions per second. If the mass of the target is 500 g and specific heat `0.09 kcal g^(-1).^@C^(-1)`, find the time taken for the temperature of the target to rise by `100^@C`, assuming the whole energy of the ions is converted to heat and absorbed by the target.

Answer

To solve the problem, we need to calculate the time taken for the temperature of the target to rise by \(100^\circ C\) when hydrogen ions strike it. Here’s the step-by-step solution:

### Step 1: Calculate the energy per ion
The kinetic energy (\(E_k\)) of a single hydrogen ion can be calculated using the formula:
\[ E_k = \frac{1}{2} mv^2 \]
where:
- \(m = 1.67 \times 10^{-27} \, \text{kg}\) (mass of a hydrogen ion)
- \(v = 2 \times 10^7 \, \text{m/s}\) (velocity of the ion)

Calculating \(E_k\):
\[ E_k = \frac{1}{2} \times (1.67 \times 10^{-27}) \times (2 \times 10^7)^2 \]
\[ E_k = \frac{1}{2} \times (1.67 \times 10^{-27}) \times (4 \times 10^{14}) \]
\[ E_k = 3.34 \times 10^{-13} \, \text{J} \]

### Step 2: Calculate the total energy per second
The total energy (\(E_{total}\)) delivered by \(10^{15}\) ions per second is:
\[ E_{total} = E_k \times \text{number of ions per second} \]
\[ E_{total} = (3.34 \times 10^{-13} \, \text{J}) \times (10^{15}) \]
\[ E_{total} = 3.34 \times 10^{2} \, \text{J/s} \]

### Step 3: Calculate the heat required to raise the temperature of the target
The heat (\(Q\)) required to raise the temperature of the target can be calculated using:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \(m = 500 \, \text{g} = 0.5 \, \text{kg}\)
- \(s = 0.09 \, \text{kcal/g}^\circ C = 0.09 \times 4184 \, \text{J/kg}^\circ C = 376.56 \, \text{J/kg}^\circ C\)
- \(\Delta T = 100^\circ C\)

Calculating \(Q\):
\[ Q = 0.5 \times 376.56 \times 100 \]
\[ Q = 18828 \, \text{J} \]

### Step 4: Calculate the time taken to raise the temperature
Using the power delivered to the target, we can find the time (\(t\)) taken to raise the temperature:
\[ t = \frac{Q}{E_{total}} \]
\[ t = \frac{18828}{334} \]
\[ t \approx 56.4 \, \text{s} \]

### Final Answer
The time taken for the temperature of the target to rise by \(100^\circ C\) is approximately \(56.4\) seconds.

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