142581
If \(\lambda_0\) is the de-Broglie wavelength for a proton accelerated through a potential difference of \(100 \mathrm{~V}\), the de-Broglie wavelength for \(\alpha\)-particle accelerated through the same potential difference is
1 \(2 \sqrt{2 \lambda_0}\)
2 \(\frac{\lambda_{\mathrm{o}}}{2}\)
3 \(\frac{\lambda_{\mathrm{o}}}{2 \sqrt{2}}\)
4 \(\frac{\lambda_{\mathrm{o}}}{\sqrt{2}}\)
Explanation:
C We know that, de - Broglie Wavelength \((\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}\) \(\lambda \propto \frac{1}{\sqrt{\mathrm{mq}}}\) \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_\alpha \cdot \mathrm{q}_\alpha}}\)( \(\because\) voltage are same) \(\because\) Mass of \(\alpha\) particle \(=\) four times of proton And Charge of \(\alpha\) particle \(=\) Twice of proton \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times \mathrm{e}}{4 \times 2 \mathrm{e}}}\) \(\lambda_\alpha=\frac{\lambda_{\mathrm{p}}}{2 \sqrt{2}}=\frac{\lambda_0}{2 \sqrt{2}}\)
EAMCET-2002
Dual nature of radiation and Matter
142466
If the kinetic energy of a moving particle is $\mathrm{E}$, then the de-Broglie wavelength is
C We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mE}})$ Therefore, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where $\lambda$ De-Broglie wavelength
VITEEE-2017
Dual nature of radiation and Matter
142415
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600 \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3} \mathrm{~W}$ will be $\left(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
1 $10^{18}$
2 $10^{17}$
3 $10^{16}$
4 $10^{15}$
Explanation:
C Given, Power $(\mathrm{P})=3.3 \times 10^{-3} \mathrm{~W}, \lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}$ $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ Number of photons $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}$ $\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}} =10^{16} \text { photons per sec }$
NEET - 2021
Dual nature of radiation and Matter
142422
An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{P}$ to $\lambda_{e}$ is $\left[m_{e}=\right.$ mass of electron, $m_{p}=$ mass of proton]
B The kinetic energy is same for both particle Momentum of electron $\mathrm{p}_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}}$ Momentum of proton $\mathrm{p}_{\mathrm{p}}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{p}}}, \lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{c}}}=\frac{\mathrm{p}_{\mathrm{c}}}{\mathrm{p}_{\mathrm{p}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{e}}}}{\sqrt{\mathrm{m}_{\mathrm{p}}}}=\left(\frac{\mathrm{m}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{p}}}\right)^{\frac{1}{2}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142581
If \(\lambda_0\) is the de-Broglie wavelength for a proton accelerated through a potential difference of \(100 \mathrm{~V}\), the de-Broglie wavelength for \(\alpha\)-particle accelerated through the same potential difference is
1 \(2 \sqrt{2 \lambda_0}\)
2 \(\frac{\lambda_{\mathrm{o}}}{2}\)
3 \(\frac{\lambda_{\mathrm{o}}}{2 \sqrt{2}}\)
4 \(\frac{\lambda_{\mathrm{o}}}{\sqrt{2}}\)
Explanation:
C We know that, de - Broglie Wavelength \((\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}\) \(\lambda \propto \frac{1}{\sqrt{\mathrm{mq}}}\) \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_\alpha \cdot \mathrm{q}_\alpha}}\)( \(\because\) voltage are same) \(\because\) Mass of \(\alpha\) particle \(=\) four times of proton And Charge of \(\alpha\) particle \(=\) Twice of proton \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times \mathrm{e}}{4 \times 2 \mathrm{e}}}\) \(\lambda_\alpha=\frac{\lambda_{\mathrm{p}}}{2 \sqrt{2}}=\frac{\lambda_0}{2 \sqrt{2}}\)
EAMCET-2002
Dual nature of radiation and Matter
142466
If the kinetic energy of a moving particle is $\mathrm{E}$, then the de-Broglie wavelength is
C We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mE}})$ Therefore, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where $\lambda$ De-Broglie wavelength
VITEEE-2017
Dual nature of radiation and Matter
142415
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600 \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3} \mathrm{~W}$ will be $\left(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
1 $10^{18}$
2 $10^{17}$
3 $10^{16}$
4 $10^{15}$
Explanation:
C Given, Power $(\mathrm{P})=3.3 \times 10^{-3} \mathrm{~W}, \lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}$ $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ Number of photons $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}$ $\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}} =10^{16} \text { photons per sec }$
NEET - 2021
Dual nature of radiation and Matter
142422
An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{P}$ to $\lambda_{e}$ is $\left[m_{e}=\right.$ mass of electron, $m_{p}=$ mass of proton]
B The kinetic energy is same for both particle Momentum of electron $\mathrm{p}_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}}$ Momentum of proton $\mathrm{p}_{\mathrm{p}}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{p}}}, \lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{c}}}=\frac{\mathrm{p}_{\mathrm{c}}}{\mathrm{p}_{\mathrm{p}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{e}}}}{\sqrt{\mathrm{m}_{\mathrm{p}}}}=\left(\frac{\mathrm{m}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{p}}}\right)^{\frac{1}{2}}$
142581
If \(\lambda_0\) is the de-Broglie wavelength for a proton accelerated through a potential difference of \(100 \mathrm{~V}\), the de-Broglie wavelength for \(\alpha\)-particle accelerated through the same potential difference is
1 \(2 \sqrt{2 \lambda_0}\)
2 \(\frac{\lambda_{\mathrm{o}}}{2}\)
3 \(\frac{\lambda_{\mathrm{o}}}{2 \sqrt{2}}\)
4 \(\frac{\lambda_{\mathrm{o}}}{\sqrt{2}}\)
Explanation:
C We know that, de - Broglie Wavelength \((\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}\) \(\lambda \propto \frac{1}{\sqrt{\mathrm{mq}}}\) \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_\alpha \cdot \mathrm{q}_\alpha}}\)( \(\because\) voltage are same) \(\because\) Mass of \(\alpha\) particle \(=\) four times of proton And Charge of \(\alpha\) particle \(=\) Twice of proton \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times \mathrm{e}}{4 \times 2 \mathrm{e}}}\) \(\lambda_\alpha=\frac{\lambda_{\mathrm{p}}}{2 \sqrt{2}}=\frac{\lambda_0}{2 \sqrt{2}}\)
EAMCET-2002
Dual nature of radiation and Matter
142466
If the kinetic energy of a moving particle is $\mathrm{E}$, then the de-Broglie wavelength is
C We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mE}})$ Therefore, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where $\lambda$ De-Broglie wavelength
VITEEE-2017
Dual nature of radiation and Matter
142415
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600 \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3} \mathrm{~W}$ will be $\left(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
1 $10^{18}$
2 $10^{17}$
3 $10^{16}$
4 $10^{15}$
Explanation:
C Given, Power $(\mathrm{P})=3.3 \times 10^{-3} \mathrm{~W}, \lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}$ $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ Number of photons $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}$ $\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}} =10^{16} \text { photons per sec }$
NEET - 2021
Dual nature of radiation and Matter
142422
An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{P}$ to $\lambda_{e}$ is $\left[m_{e}=\right.$ mass of electron, $m_{p}=$ mass of proton]
B The kinetic energy is same for both particle Momentum of electron $\mathrm{p}_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}}$ Momentum of proton $\mathrm{p}_{\mathrm{p}}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{p}}}, \lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{c}}}=\frac{\mathrm{p}_{\mathrm{c}}}{\mathrm{p}_{\mathrm{p}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{e}}}}{\sqrt{\mathrm{m}_{\mathrm{p}}}}=\left(\frac{\mathrm{m}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{p}}}\right)^{\frac{1}{2}}$
142581
If \(\lambda_0\) is the de-Broglie wavelength for a proton accelerated through a potential difference of \(100 \mathrm{~V}\), the de-Broglie wavelength for \(\alpha\)-particle accelerated through the same potential difference is
1 \(2 \sqrt{2 \lambda_0}\)
2 \(\frac{\lambda_{\mathrm{o}}}{2}\)
3 \(\frac{\lambda_{\mathrm{o}}}{2 \sqrt{2}}\)
4 \(\frac{\lambda_{\mathrm{o}}}{\sqrt{2}}\)
Explanation:
C We know that, de - Broglie Wavelength \((\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}\) \(\lambda \propto \frac{1}{\sqrt{\mathrm{mq}}}\) \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_\alpha \cdot \mathrm{q}_\alpha}}\)( \(\because\) voltage are same) \(\because\) Mass of \(\alpha\) particle \(=\) four times of proton And Charge of \(\alpha\) particle \(=\) Twice of proton \(\frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times \mathrm{e}}{4 \times 2 \mathrm{e}}}\) \(\lambda_\alpha=\frac{\lambda_{\mathrm{p}}}{2 \sqrt{2}}=\frac{\lambda_0}{2 \sqrt{2}}\)
EAMCET-2002
Dual nature of radiation and Matter
142466
If the kinetic energy of a moving particle is $\mathrm{E}$, then the de-Broglie wavelength is
C We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mE}})$ Therefore, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where $\lambda$ De-Broglie wavelength
VITEEE-2017
Dual nature of radiation and Matter
142415
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600 \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3} \mathrm{~W}$ will be $\left(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
1 $10^{18}$
2 $10^{17}$
3 $10^{16}$
4 $10^{15}$
Explanation:
C Given, Power $(\mathrm{P})=3.3 \times 10^{-3} \mathrm{~W}, \lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}$ $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ Number of photons $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}$ $\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}} =10^{16} \text { photons per sec }$
NEET - 2021
Dual nature of radiation and Matter
142422
An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{P}$ to $\lambda_{e}$ is $\left[m_{e}=\right.$ mass of electron, $m_{p}=$ mass of proton]
B The kinetic energy is same for both particle Momentum of electron $\mathrm{p}_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}}$ Momentum of proton $\mathrm{p}_{\mathrm{p}}=\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{p}}}, \lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{c}}}=\frac{\mathrm{p}_{\mathrm{c}}}{\mathrm{p}_{\mathrm{p}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{e}}}}{\sqrt{\mathrm{m}_{\mathrm{p}}}}=\left(\frac{\mathrm{m}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{p}}}\right)^{\frac{1}{2}}$