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A proton a deutron and an alpha particle...

A proton a deutron and an `alpha` particle are accelerated through potentials of `V, 2V` and `4V` respectively. Their velocity will bear a ratio

A

`1:1:1`

B

`1:sqrt2:1`

C

`sqrt2:1:1`

D

`1:1:sqrt2`

Text Solution

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The correct Answer is:
To find the ratio of velocities of a proton, a deuteron, and an alpha particle when they are accelerated through potentials of \( V \), \( 2V \), and \( 4V \) respectively, we can use the principle of energy conservation. The kinetic energy gained by a charged particle when accelerated through a potential difference is given by: \[ \text{K.E.} = QV \] Where \( Q \) is the charge of the particle and \( V \) is the potential difference. The kinetic energy can also be expressed in terms of mass and velocity: \[ \text{K.E.} = \frac{1}{2} mv^2 \] Equating the two expressions for kinetic energy, we have: \[ \frac{1}{2} mv^2 = QV \] From this, we can solve for velocity \( v \): \[ v = \sqrt{\frac{2QV}{m}} \] ### Step 1: Calculate the velocity of the proton For the proton: - Charge \( Q_p = e \) - Mass \( m_p = m \) - Potential \( V_p = V \) Thus, the velocity \( v_p \) is: \[ v_p = \sqrt{\frac{2eV}{m}} \] ### Step 2: Calculate the velocity of the deuteron For the deuteron: - Charge \( Q_d = e \) - Mass \( m_d = 2m \) - Potential \( V_d = 2V \) Thus, the velocity \( v_d \) is: \[ v_d = \sqrt{\frac{2e(2V)}{2m}} = \sqrt{\frac{4eV}{2m}} = \sqrt{\frac{2eV}{m}} = v_p \] ### Step 3: Calculate the velocity of the alpha particle For the alpha particle: - Charge \( Q_{\alpha} = 2e \) - Mass \( m_{\alpha} = 4m \) - Potential \( V_{\alpha} = 4V \) Thus, the velocity \( v_{\alpha} \) is: \[ v_{\alpha} = \sqrt{\frac{2(2e)(4V)}{4m}} = \sqrt{\frac{16eV}{4m}} = \sqrt{\frac{4eV}{m}} = 2\sqrt{\frac{eV}{m}} = 2v_p \] ### Step 4: Determine the ratio of velocities Now we have: - \( v_p = \sqrt{\frac{2eV}{m}} \) - \( v_d = v_p \) - \( v_{\alpha} = 2v_p \) The ratio of velocities \( v_p : v_d : v_{\alpha} \) is: \[ 1 : 1 : 2 \] ### Final Answer The ratio of the velocities of the proton, deuteron, and alpha particle is: \[ 1 : 1 : 2 \]

To find the ratio of velocities of a proton, a deuteron, and an alpha particle when they are accelerated through potentials of \( V \), \( 2V \), and \( 4V \) respectively, we can use the principle of energy conservation. The kinetic energy gained by a charged particle when accelerated through a potential difference is given by: \[ \text{K.E.} = QV \] Where \( Q \) is the charge of the particle and \( V \) is the potential difference. The kinetic energy can also be expressed in terms of mass and velocity: ...
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Knowledge Check

  • A protons , a deuteron and an alpha-particle are accelerated through the same p.d of V volt. The velocities acquired by them are in the ratio

    A
    `1:1:sqrt2`
    B
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    B
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    C
    `2 sqrt(2) : 1`
    D
    `(1)/( 2 sqrt(2)) : 1`
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    A
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