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The de-Broglie wavelength of a proton and \(\alpha\) - particle is same. The ratio of their velocities will be.......... ( \(\alpha\) particle is the He- nucleus, having two protons and two neutrons. Thus, its mass \(\mathrm{M}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\) where \(\mathrm{m}_{\mathrm{p}}\) is the mass of the proton.) (A) \(1: 4\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)

Short Answer

Expert verified
The ratio of their velocities is \(2:1\), which corresponds to choice (C).

Step by step solution

01

Write down the de-Broglie wavelength formula for both particles.

For a particle with mass m and velocity v, the de-Broglie wavelength is given by the formula, \(\lambda = \frac{h}{mv}\), where h is Planck's constant. For a proton, with mass \(m_p\) and velocity \(v_p\), its de-Broglie wavelength \(\lambda_p = \frac{h}{m_p v_p}\). For an \(\alpha\)-particle, with mass \(M_\alpha = 4m_p\) and velocity \(v_\alpha\), its de-Broglie wavelength \(\lambda_\alpha = \frac{h}{(4m_p) v_\alpha}\).
02

Set up the equation with the given condition and solve for velocity ratio.

We are given that the de-Broglie wavelengths of the proton and the \(\alpha\)-particle are the same, i.e., \(\lambda_p = \lambda_\alpha\). Substitute the formulas from Step 1: \(\frac{h}{m_pv_p} = \frac{h}{(4m_p)v_\alpha}\) Now, we need to find the ratio of the velocities, \(\frac{v_p}{v_\alpha}\). Divide both sides by \(h\) and rearrange the equation: \(\frac{v_p}{v_\alpha} = \frac{4m_p v_\alpha}{m_pv_p}\) To solve for the velocity ratio, multiply both sides by \(v_p\) and divide both sides by \(4m_pv_\alpha\): \(\frac{v_p}{v_\alpha} \times \frac{v_p}{4m_p} \times \frac{1}{v_\alpha} = \frac{4m_p v_\alpha}{m_p v_p} \times \frac{v_p}{4m_p} \times \frac{1}{v_\alpha}\) This simplifies to: \(\frac{v_p^2}{4v_\alpha^2} = 1\) Take the square-root of both sides: \(\frac{v_p}{2v_\alpha} = 1\)
03

Find the required ratio of velocities.

We have found that \(\frac{v_p}{2v_\alpha} = 1\), which implies: \(v_p = 2v_\alpha\) Now, divide both sides by \(v_\alpha\): \(\frac{v_p}{v_\alpha} = 2\) So, the ratio of their velocities is \(2:1\), which corresponds to choice (C).

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Most popular questions from this chapter

Energy of photon having wavelength \(\lambda\) is \(2 \mathrm{eV}\). This photon when incident on metal. Maximum velocity of emitted is \(\mathrm{V}\). If \(\lambda\) is decreased \(25 \%\) and maximum velocity is made double, work function of metal is \(\ldots \ldots \ldots . . \mathrm{V}\) (A) \(1.2\) (B) \(1.5\) (C) \(1.6\) (D) \(1.8\)

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)

Work function of \(\mathrm{Zn}\) is \(3.74 \mathrm{eV}\). If the sphere of \(\mathrm{Zn}\) is illuminated by the X-ray of wavelength \(12 \AA\) the maximum potential produced on the sphere is \(\ldots \ldots \ldots\) \(\left(\mathrm{h}-6.625 \times 10^{-34 \mathrm{~J}-\mathrm{s}}\right)\) (A) \(10.314 \mathrm{~V}\) (B) \(103.14 \mathrm{~V}\) (C) \(1031.4 \mathrm{~V}\) (D) \(10314 \mathrm{~V}\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (A) Planck's theory of quantum (p) Light energy \(=\mathrm{hv}\) (B) Einstein's theory of quanta (q) Angular momentum of electron in an orbit. (C) Bohr's stationary orbit (r) Oscillator energies (D) D-Broglie waves (s) Electron microscope (A) \((\mathrm{A}-\mathrm{p}),(\mathrm{b}-\mathrm{q}),(\mathrm{C}-\mathrm{r}),(\mathrm{D}-\mathrm{s})\) (B) \((\mathrm{A}-\mathrm{q}),(\mathrm{B}-\mathrm{r}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{p})\) (C) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{q}),(\mathrm{D}-\mathrm{s})\) (D) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{q})\)

A body of mass \(200 \mathrm{~g}\) moves at the speed of $5 \mathrm{~m} / \mathrm{hr}$. So deBroglie wavelength related to it is of the order........ \(\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (A) \(10^{-10} \mathrm{~m}\) (B) \(10^{-20} \mathrm{~m}\) (C) \(10^{-30} \mathrm{~m}\) (D) \(10^{-40} \mathrm{~m}\)

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