142530
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{~kg}$, charge $=1.6 \times 10^{-19} \mathrm{C}$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck' constant $=$ $6.6 \times 10^{-34} \mathrm{Js}$
142531
The de-Broglie wavelength of an electron moving with a velocity $\mathrm{c} / 2(\mathrm{c}=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is
142532
A beam of charged particles, having been accelerated by a voltage $V$, has a wavelength $\lambda$. On increasing the accelerating voltage to $4 \mathrm{~V}$, the wavelength will become
1 $2 \lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 remain the same
Explanation:
B We know, de-Broglie wavelength associated with energy- $\text { K.E. }=E=\frac{p^{2}}{2 m}$ $p=\sqrt{2 m E}$ $\mathrm{E} \rightarrow$ Kinetic energy of the particles $\mathrm{m} \rightarrow$ Mass of the particles Then, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ We know that, $\mathrm{E}=\mathrm{eV}$ $\mathrm{E} \propto \mathrm{V}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\therefore \quad \frac{\lambda^{\prime}}{\lambda} =\sqrt{\frac{\mathrm{V}}{\mathrm{V}^{\prime}}}$ $\because \quad \mathrm{V}^{\prime} =4 \mathrm{~V}$ $\text { So, } \lambda^{\prime}=\lambda \cdot \sqrt{\frac{\mathrm{V}}{4 \mathrm{~V}}}$ $\lambda^{\prime} =\lambda \cdot \frac{1}{\sqrt{4}}=\frac{\lambda}{2}$ $\lambda^{\prime} =\frac{\lambda}{2}$
142530
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{~kg}$, charge $=1.6 \times 10^{-19} \mathrm{C}$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck' constant $=$ $6.6 \times 10^{-34} \mathrm{Js}$
142531
The de-Broglie wavelength of an electron moving with a velocity $\mathrm{c} / 2(\mathrm{c}=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is
142532
A beam of charged particles, having been accelerated by a voltage $V$, has a wavelength $\lambda$. On increasing the accelerating voltage to $4 \mathrm{~V}$, the wavelength will become
1 $2 \lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 remain the same
Explanation:
B We know, de-Broglie wavelength associated with energy- $\text { K.E. }=E=\frac{p^{2}}{2 m}$ $p=\sqrt{2 m E}$ $\mathrm{E} \rightarrow$ Kinetic energy of the particles $\mathrm{m} \rightarrow$ Mass of the particles Then, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ We know that, $\mathrm{E}=\mathrm{eV}$ $\mathrm{E} \propto \mathrm{V}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\therefore \quad \frac{\lambda^{\prime}}{\lambda} =\sqrt{\frac{\mathrm{V}}{\mathrm{V}^{\prime}}}$ $\because \quad \mathrm{V}^{\prime} =4 \mathrm{~V}$ $\text { So, } \lambda^{\prime}=\lambda \cdot \sqrt{\frac{\mathrm{V}}{4 \mathrm{~V}}}$ $\lambda^{\prime} =\lambda \cdot \frac{1}{\sqrt{4}}=\frac{\lambda}{2}$ $\lambda^{\prime} =\frac{\lambda}{2}$
142530
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{~kg}$, charge $=1.6 \times 10^{-19} \mathrm{C}$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck' constant $=$ $6.6 \times 10^{-34} \mathrm{Js}$
142531
The de-Broglie wavelength of an electron moving with a velocity $\mathrm{c} / 2(\mathrm{c}=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is
142532
A beam of charged particles, having been accelerated by a voltage $V$, has a wavelength $\lambda$. On increasing the accelerating voltage to $4 \mathrm{~V}$, the wavelength will become
1 $2 \lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 remain the same
Explanation:
B We know, de-Broglie wavelength associated with energy- $\text { K.E. }=E=\frac{p^{2}}{2 m}$ $p=\sqrt{2 m E}$ $\mathrm{E} \rightarrow$ Kinetic energy of the particles $\mathrm{m} \rightarrow$ Mass of the particles Then, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ We know that, $\mathrm{E}=\mathrm{eV}$ $\mathrm{E} \propto \mathrm{V}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\therefore \quad \frac{\lambda^{\prime}}{\lambda} =\sqrt{\frac{\mathrm{V}}{\mathrm{V}^{\prime}}}$ $\because \quad \mathrm{V}^{\prime} =4 \mathrm{~V}$ $\text { So, } \lambda^{\prime}=\lambda \cdot \sqrt{\frac{\mathrm{V}}{4 \mathrm{~V}}}$ $\lambda^{\prime} =\lambda \cdot \frac{1}{\sqrt{4}}=\frac{\lambda}{2}$ $\lambda^{\prime} =\frac{\lambda}{2}$
142530
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{~kg}$, charge $=1.6 \times 10^{-19} \mathrm{C}$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck' constant $=$ $6.6 \times 10^{-34} \mathrm{Js}$
142531
The de-Broglie wavelength of an electron moving with a velocity $\mathrm{c} / 2(\mathrm{c}=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is
142532
A beam of charged particles, having been accelerated by a voltage $V$, has a wavelength $\lambda$. On increasing the accelerating voltage to $4 \mathrm{~V}$, the wavelength will become
1 $2 \lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 remain the same
Explanation:
B We know, de-Broglie wavelength associated with energy- $\text { K.E. }=E=\frac{p^{2}}{2 m}$ $p=\sqrt{2 m E}$ $\mathrm{E} \rightarrow$ Kinetic energy of the particles $\mathrm{m} \rightarrow$ Mass of the particles Then, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ We know that, $\mathrm{E}=\mathrm{eV}$ $\mathrm{E} \propto \mathrm{V}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\therefore \quad \frac{\lambda^{\prime}}{\lambda} =\sqrt{\frac{\mathrm{V}}{\mathrm{V}^{\prime}}}$ $\because \quad \mathrm{V}^{\prime} =4 \mathrm{~V}$ $\text { So, } \lambda^{\prime}=\lambda \cdot \sqrt{\frac{\mathrm{V}}{4 \mathrm{~V}}}$ $\lambda^{\prime} =\lambda \cdot \frac{1}{\sqrt{4}}=\frac{\lambda}{2}$ $\lambda^{\prime} =\frac{\lambda}{2}$
