142413
If the particles listed below all have the same kinetic energy, which one would possess the shortest de Broglie wavelength?
1 Deuteron
2 $\alpha$-particle
3 Proton
4 Electron
Explanation:
B We know, the de- Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}$ Where, $\mathrm{m}$ is mass and $\mathrm{k}$ is kinetic energy. All charged particle have same kinetic energy (Given) $\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ As, $\mathrm{m}_{\text {electron }} \lt \mathrm{m}_{\text {proton }} \lt \mathrm{m}_{\text {deutron }} \lt \mathrm{m}_{\alpha \text {-particle }}$ $\therefore \lambda_{\text {electron }}>\lambda_{\text {proton }}>\lambda_{\text {deutron }}>\lambda_{\alpha-\text { particle }}$ Hence, $\alpha$-particle has the shortest de-Broglie wavelength.
Shift-I
Dual nature of radiation and Matter
142414
A proton and an alpha particle are accelerated through the same potential difference. The ratio of the wavelengths associated with proton and alpha particle respectively is
1 $1: 2 \sqrt{2}$
2 $2: 1$
3 $2 \sqrt{2}: 1$
4 $4: 1$
Explanation:
C Let the mass of the proton be $\mathrm{m}_{\mathrm{p}}$ and the momentum of the proton is given by- $\mathrm{p}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ The de - Broglie wavelength will be- $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}} \quad\left(\because \mathrm{K}=\mathrm{q}_{\mathrm{p}} \mathrm{V}\right)$ Therefore, $\lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}$ Similarly, For alpha particle - $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}$ $\left(\because \mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}, \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ Now, the ratio is - $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}}{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}}$ $\frac{\lambda_{\mathrm{P}}}{\lambda_{\alpha}}=\frac{2 \sqrt{2}}{1}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=2 \sqrt{2}: 1$
2012
Dual nature of radiation and Matter
142416
The de Broglie wavelength associated with a proton under the influence of an electric potential of 100 volts is
1 $1.227 \AA$
2 $2.86 \mathrm{pm}$
3 $12.27 \AA$
4 $1.146 \times 10^{-21} \mathrm{~m}$
Explanation:
B Given, potential (V) $=100 \mathrm{~V}$ Mass of proton $(\mathrm{m})=1.672 \times 10^{-27} \mathrm{~kg}$ Charge of proton $(q)=1.6 \times 10^{-19} \mathrm{C}$ We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2\left(1.672 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 100}}$ $\lambda=2.853 \times 10^{-12} \mathrm{~m}$ $\lambda=2.86 \mathrm{pm}$
AP EAMCET-19.08.2021
Dual nature of radiation and Matter
142417
The de-Broglie wavelength of an electron having kinetic energy $100 \mathrm{eV}$ is [Use $\mathrm{h}=4.14 \times$ $10^{-15} \mathrm{eV}$, mass of electron $=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}} \mathrm{eV}, 1 \mathrm{pm}=$ $10^{-12} \mathrm{ml}$
1 $150.1 \mathrm{pm}$
2 $124.2 \mathrm{pm}$
3 $115.5 \mathrm{pm}$
4 $120.8 \mathrm{pm}$
Explanation:
B Given Kinetic Energy $=100 \mathrm{eV}$ $\mathrm{h}=4.14 \times 10^{-15} \mathrm{eV}$ mass of electron, $\mathrm{m}=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}}$ $1 \mathrm{pm} .=10^{-12} \mathrm{~m}$. We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}$ $\lambda=\frac{4.14 \times 10^{-15} \times \mathrm{c}}{\sqrt{2 \times 0.5 \times 10^{6} \times 100}}$ $\lambda=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{\sqrt{10^{8}}}$ $\lambda=12.42 \times 10^{-11} \mathrm{~m}$ $\lambda=124.2 \times 10^{-12} \mathrm{~m}$ $\lambda=124.2 \mathrm{pm}$
142413
If the particles listed below all have the same kinetic energy, which one would possess the shortest de Broglie wavelength?
1 Deuteron
2 $\alpha$-particle
3 Proton
4 Electron
Explanation:
B We know, the de- Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}$ Where, $\mathrm{m}$ is mass and $\mathrm{k}$ is kinetic energy. All charged particle have same kinetic energy (Given) $\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ As, $\mathrm{m}_{\text {electron }} \lt \mathrm{m}_{\text {proton }} \lt \mathrm{m}_{\text {deutron }} \lt \mathrm{m}_{\alpha \text {-particle }}$ $\therefore \lambda_{\text {electron }}>\lambda_{\text {proton }}>\lambda_{\text {deutron }}>\lambda_{\alpha-\text { particle }}$ Hence, $\alpha$-particle has the shortest de-Broglie wavelength.
Shift-I
Dual nature of radiation and Matter
142414
A proton and an alpha particle are accelerated through the same potential difference. The ratio of the wavelengths associated with proton and alpha particle respectively is
1 $1: 2 \sqrt{2}$
2 $2: 1$
3 $2 \sqrt{2}: 1$
4 $4: 1$
Explanation:
C Let the mass of the proton be $\mathrm{m}_{\mathrm{p}}$ and the momentum of the proton is given by- $\mathrm{p}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ The de - Broglie wavelength will be- $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}} \quad\left(\because \mathrm{K}=\mathrm{q}_{\mathrm{p}} \mathrm{V}\right)$ Therefore, $\lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}$ Similarly, For alpha particle - $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}$ $\left(\because \mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}, \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ Now, the ratio is - $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}}{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}}$ $\frac{\lambda_{\mathrm{P}}}{\lambda_{\alpha}}=\frac{2 \sqrt{2}}{1}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=2 \sqrt{2}: 1$
2012
Dual nature of radiation and Matter
142416
The de Broglie wavelength associated with a proton under the influence of an electric potential of 100 volts is
1 $1.227 \AA$
2 $2.86 \mathrm{pm}$
3 $12.27 \AA$
4 $1.146 \times 10^{-21} \mathrm{~m}$
Explanation:
B Given, potential (V) $=100 \mathrm{~V}$ Mass of proton $(\mathrm{m})=1.672 \times 10^{-27} \mathrm{~kg}$ Charge of proton $(q)=1.6 \times 10^{-19} \mathrm{C}$ We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2\left(1.672 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 100}}$ $\lambda=2.853 \times 10^{-12} \mathrm{~m}$ $\lambda=2.86 \mathrm{pm}$
AP EAMCET-19.08.2021
Dual nature of radiation and Matter
142417
The de-Broglie wavelength of an electron having kinetic energy $100 \mathrm{eV}$ is [Use $\mathrm{h}=4.14 \times$ $10^{-15} \mathrm{eV}$, mass of electron $=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}} \mathrm{eV}, 1 \mathrm{pm}=$ $10^{-12} \mathrm{ml}$
1 $150.1 \mathrm{pm}$
2 $124.2 \mathrm{pm}$
3 $115.5 \mathrm{pm}$
4 $120.8 \mathrm{pm}$
Explanation:
B Given Kinetic Energy $=100 \mathrm{eV}$ $\mathrm{h}=4.14 \times 10^{-15} \mathrm{eV}$ mass of electron, $\mathrm{m}=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}}$ $1 \mathrm{pm} .=10^{-12} \mathrm{~m}$. We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}$ $\lambda=\frac{4.14 \times 10^{-15} \times \mathrm{c}}{\sqrt{2 \times 0.5 \times 10^{6} \times 100}}$ $\lambda=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{\sqrt{10^{8}}}$ $\lambda=12.42 \times 10^{-11} \mathrm{~m}$ $\lambda=124.2 \times 10^{-12} \mathrm{~m}$ $\lambda=124.2 \mathrm{pm}$
142413
If the particles listed below all have the same kinetic energy, which one would possess the shortest de Broglie wavelength?
1 Deuteron
2 $\alpha$-particle
3 Proton
4 Electron
Explanation:
B We know, the de- Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}$ Where, $\mathrm{m}$ is mass and $\mathrm{k}$ is kinetic energy. All charged particle have same kinetic energy (Given) $\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ As, $\mathrm{m}_{\text {electron }} \lt \mathrm{m}_{\text {proton }} \lt \mathrm{m}_{\text {deutron }} \lt \mathrm{m}_{\alpha \text {-particle }}$ $\therefore \lambda_{\text {electron }}>\lambda_{\text {proton }}>\lambda_{\text {deutron }}>\lambda_{\alpha-\text { particle }}$ Hence, $\alpha$-particle has the shortest de-Broglie wavelength.
Shift-I
Dual nature of radiation and Matter
142414
A proton and an alpha particle are accelerated through the same potential difference. The ratio of the wavelengths associated with proton and alpha particle respectively is
1 $1: 2 \sqrt{2}$
2 $2: 1$
3 $2 \sqrt{2}: 1$
4 $4: 1$
Explanation:
C Let the mass of the proton be $\mathrm{m}_{\mathrm{p}}$ and the momentum of the proton is given by- $\mathrm{p}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ The de - Broglie wavelength will be- $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}} \quad\left(\because \mathrm{K}=\mathrm{q}_{\mathrm{p}} \mathrm{V}\right)$ Therefore, $\lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}$ Similarly, For alpha particle - $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}$ $\left(\because \mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}, \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ Now, the ratio is - $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}}{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}}$ $\frac{\lambda_{\mathrm{P}}}{\lambda_{\alpha}}=\frac{2 \sqrt{2}}{1}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=2 \sqrt{2}: 1$
2012
Dual nature of radiation and Matter
142416
The de Broglie wavelength associated with a proton under the influence of an electric potential of 100 volts is
1 $1.227 \AA$
2 $2.86 \mathrm{pm}$
3 $12.27 \AA$
4 $1.146 \times 10^{-21} \mathrm{~m}$
Explanation:
B Given, potential (V) $=100 \mathrm{~V}$ Mass of proton $(\mathrm{m})=1.672 \times 10^{-27} \mathrm{~kg}$ Charge of proton $(q)=1.6 \times 10^{-19} \mathrm{C}$ We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2\left(1.672 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 100}}$ $\lambda=2.853 \times 10^{-12} \mathrm{~m}$ $\lambda=2.86 \mathrm{pm}$
AP EAMCET-19.08.2021
Dual nature of radiation and Matter
142417
The de-Broglie wavelength of an electron having kinetic energy $100 \mathrm{eV}$ is [Use $\mathrm{h}=4.14 \times$ $10^{-15} \mathrm{eV}$, mass of electron $=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}} \mathrm{eV}, 1 \mathrm{pm}=$ $10^{-12} \mathrm{ml}$
1 $150.1 \mathrm{pm}$
2 $124.2 \mathrm{pm}$
3 $115.5 \mathrm{pm}$
4 $120.8 \mathrm{pm}$
Explanation:
B Given Kinetic Energy $=100 \mathrm{eV}$ $\mathrm{h}=4.14 \times 10^{-15} \mathrm{eV}$ mass of electron, $\mathrm{m}=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}}$ $1 \mathrm{pm} .=10^{-12} \mathrm{~m}$. We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}$ $\lambda=\frac{4.14 \times 10^{-15} \times \mathrm{c}}{\sqrt{2 \times 0.5 \times 10^{6} \times 100}}$ $\lambda=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{\sqrt{10^{8}}}$ $\lambda=12.42 \times 10^{-11} \mathrm{~m}$ $\lambda=124.2 \times 10^{-12} \mathrm{~m}$ $\lambda=124.2 \mathrm{pm}$
142413
If the particles listed below all have the same kinetic energy, which one would possess the shortest de Broglie wavelength?
1 Deuteron
2 $\alpha$-particle
3 Proton
4 Electron
Explanation:
B We know, the de- Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}$ Where, $\mathrm{m}$ is mass and $\mathrm{k}$ is kinetic energy. All charged particle have same kinetic energy (Given) $\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ As, $\mathrm{m}_{\text {electron }} \lt \mathrm{m}_{\text {proton }} \lt \mathrm{m}_{\text {deutron }} \lt \mathrm{m}_{\alpha \text {-particle }}$ $\therefore \lambda_{\text {electron }}>\lambda_{\text {proton }}>\lambda_{\text {deutron }}>\lambda_{\alpha-\text { particle }}$ Hence, $\alpha$-particle has the shortest de-Broglie wavelength.
Shift-I
Dual nature of radiation and Matter
142414
A proton and an alpha particle are accelerated through the same potential difference. The ratio of the wavelengths associated with proton and alpha particle respectively is
1 $1: 2 \sqrt{2}$
2 $2: 1$
3 $2 \sqrt{2}: 1$
4 $4: 1$
Explanation:
C Let the mass of the proton be $\mathrm{m}_{\mathrm{p}}$ and the momentum of the proton is given by- $\mathrm{p}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ The de - Broglie wavelength will be- $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}} \quad\left(\because \mathrm{K}=\mathrm{q}_{\mathrm{p}} \mathrm{V}\right)$ Therefore, $\lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}$ Similarly, For alpha particle - $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}$ $\left(\because \mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}, \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ Now, the ratio is - $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}}}}{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{q}_{\mathrm{p}} \times \mathrm{V}}}}$ $\frac{\lambda_{\mathrm{P}}}{\lambda_{\alpha}}=\frac{2 \sqrt{2}}{1}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=2 \sqrt{2}: 1$
2012
Dual nature of radiation and Matter
142416
The de Broglie wavelength associated with a proton under the influence of an electric potential of 100 volts is
1 $1.227 \AA$
2 $2.86 \mathrm{pm}$
3 $12.27 \AA$
4 $1.146 \times 10^{-21} \mathrm{~m}$
Explanation:
B Given, potential (V) $=100 \mathrm{~V}$ Mass of proton $(\mathrm{m})=1.672 \times 10^{-27} \mathrm{~kg}$ Charge of proton $(q)=1.6 \times 10^{-19} \mathrm{C}$ We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2\left(1.672 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 100}}$ $\lambda=2.853 \times 10^{-12} \mathrm{~m}$ $\lambda=2.86 \mathrm{pm}$
AP EAMCET-19.08.2021
Dual nature of radiation and Matter
142417
The de-Broglie wavelength of an electron having kinetic energy $100 \mathrm{eV}$ is [Use $\mathrm{h}=4.14 \times$ $10^{-15} \mathrm{eV}$, mass of electron $=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}} \mathrm{eV}, 1 \mathrm{pm}=$ $10^{-12} \mathrm{ml}$
1 $150.1 \mathrm{pm}$
2 $124.2 \mathrm{pm}$
3 $115.5 \mathrm{pm}$
4 $120.8 \mathrm{pm}$
Explanation:
B Given Kinetic Energy $=100 \mathrm{eV}$ $\mathrm{h}=4.14 \times 10^{-15} \mathrm{eV}$ mass of electron, $\mathrm{m}=\frac{0.5 \times 10^{6}}{\mathrm{c}^{2}}$ $1 \mathrm{pm} .=10^{-12} \mathrm{~m}$. We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}$ $\lambda=\frac{4.14 \times 10^{-15} \times \mathrm{c}}{\sqrt{2 \times 0.5 \times 10^{6} \times 100}}$ $\lambda=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{\sqrt{10^{8}}}$ $\lambda=12.42 \times 10^{-11} \mathrm{~m}$ $\lambda=124.2 \times 10^{-12} \mathrm{~m}$ $\lambda=124.2 \mathrm{pm}$
