NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142431
The wavelength ' $\lambda$ ' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is $(\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $h=$ Plank's constant)
1 $\frac{2 \lambda}{\mathrm{mch}}$
2 $\frac{\lambda \mathrm{mc}}{2 \mathrm{~h}}$
3 $\frac{\lambda \mathrm{m}}{4 \mathrm{~h}}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D The de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { So, } \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { Or } \quad \mathrm{v} =\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Energy of photon, $\mathrm{E}_{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}_{\mathrm{e}}=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{E}_{\mathrm{e}}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2} \mathrm{mv^{2 }}}=\frac{2 \mathrm{hc}}{\lambda \mathrm{mv}^{2}}$ Substituting the value of $\mathrm{v}$ in above equation, $\frac{E_{p}}{E_{e}}=\frac{2 h c}{\lambda m\left(\frac{h}{m \lambda}\right)^{2}}=\frac{2 \lambda m c}{h}$
MHT-CET 2020
Dual nature of radiation and Matter
142432
A photon and an electron have equal energy E. $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ is proportional to
1 $\sqrt{\mathrm{E}}$
2 $\frac{1}{\sqrt{\mathrm{E}}}$
3 $\frac{1}{\mathrm{E}}$
4 does not depend upon $\mathrm{E}$
Explanation:
B Wavelength of photon, $\lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$ Where, $\mathrm{h}=$ Planck constant $\mathrm{E}=$ Photon energy $\mathrm{c}=$ Speed of light $\lambda=$ Photon wavelength And wavelength of electron, $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Now, by equation (i) \(ii), we get- $\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} =\frac{\mathrm{hc}}{\mathrm{E}} \times \frac{\sqrt{2 \mathrm{mE}}}{\mathrm{h}}$ $=\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} \propto \frac{1}{\sqrt{\mathrm{E}}}$
Manipal UGET-2018
Dual nature of radiation and Matter
142433
An electron moving with initial velocity $\overrightarrow{\mathrm{V}}=\mathrm{V}_{0} \hat{\mathrm{i}}$ is moving in an magnetic field $\vec{B}=B_{0} \hat{J}$ then its de-Broglie wavelength
1 decreases with time
2 first increases and then decreases
3 increases with time
4 remains constant
Explanation:
D Given that, $\vec{V}=V_{0} \hat{i}$ $\vec{B}=B_{0} \hat{j}$ From de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{F} =\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=\mathrm{qVB} \hat{\mathrm{k}}$ The above force is perpendicular to $\vec{V}$ and $\vec{B}$ for the magnitude of $\mathrm{V}$ will not change. $\therefore$ Momentum (p) will not change So, $\lambda$ will also not change.
Karnataka CET-2019
Dual nature of radiation and Matter
142434
How much is the de-Broglie wavelength for an electron accelerated by an $100 \mathrm{~V}$ potential difference?
1 $0.123 \mathrm{~nm}$
2 $123 \mathrm{~nm}$
3 $12.3 \mathrm{~nm}$
4 $0.123 \mathrm{~cm}$
Explanation:
A Given that, Planck's constant, $(\mathrm{h})=6.62 \times 10^{34} \mathrm{Js}$ $\mathrm{V}=100 \text { Volt }$ Mass of electron, $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}$ $\lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$ $1=1.24 \times 10^{-10} \mathrm{~m}$ $1=0.124 \times 10^{-9} \mathrm{~m}$ $\lambda=0.12 \mathrm{~nm}$
142431
The wavelength ' $\lambda$ ' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is $(\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $h=$ Plank's constant)
1 $\frac{2 \lambda}{\mathrm{mch}}$
2 $\frac{\lambda \mathrm{mc}}{2 \mathrm{~h}}$
3 $\frac{\lambda \mathrm{m}}{4 \mathrm{~h}}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D The de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { So, } \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { Or } \quad \mathrm{v} =\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Energy of photon, $\mathrm{E}_{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}_{\mathrm{e}}=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{E}_{\mathrm{e}}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2} \mathrm{mv^{2 }}}=\frac{2 \mathrm{hc}}{\lambda \mathrm{mv}^{2}}$ Substituting the value of $\mathrm{v}$ in above equation, $\frac{E_{p}}{E_{e}}=\frac{2 h c}{\lambda m\left(\frac{h}{m \lambda}\right)^{2}}=\frac{2 \lambda m c}{h}$
MHT-CET 2020
Dual nature of radiation and Matter
142432
A photon and an electron have equal energy E. $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ is proportional to
1 $\sqrt{\mathrm{E}}$
2 $\frac{1}{\sqrt{\mathrm{E}}}$
3 $\frac{1}{\mathrm{E}}$
4 does not depend upon $\mathrm{E}$
Explanation:
B Wavelength of photon, $\lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$ Where, $\mathrm{h}=$ Planck constant $\mathrm{E}=$ Photon energy $\mathrm{c}=$ Speed of light $\lambda=$ Photon wavelength And wavelength of electron, $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Now, by equation (i) \(ii), we get- $\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} =\frac{\mathrm{hc}}{\mathrm{E}} \times \frac{\sqrt{2 \mathrm{mE}}}{\mathrm{h}}$ $=\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} \propto \frac{1}{\sqrt{\mathrm{E}}}$
Manipal UGET-2018
Dual nature of radiation and Matter
142433
An electron moving with initial velocity $\overrightarrow{\mathrm{V}}=\mathrm{V}_{0} \hat{\mathrm{i}}$ is moving in an magnetic field $\vec{B}=B_{0} \hat{J}$ then its de-Broglie wavelength
1 decreases with time
2 first increases and then decreases
3 increases with time
4 remains constant
Explanation:
D Given that, $\vec{V}=V_{0} \hat{i}$ $\vec{B}=B_{0} \hat{j}$ From de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{F} =\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=\mathrm{qVB} \hat{\mathrm{k}}$ The above force is perpendicular to $\vec{V}$ and $\vec{B}$ for the magnitude of $\mathrm{V}$ will not change. $\therefore$ Momentum (p) will not change So, $\lambda$ will also not change.
Karnataka CET-2019
Dual nature of radiation and Matter
142434
How much is the de-Broglie wavelength for an electron accelerated by an $100 \mathrm{~V}$ potential difference?
1 $0.123 \mathrm{~nm}$
2 $123 \mathrm{~nm}$
3 $12.3 \mathrm{~nm}$
4 $0.123 \mathrm{~cm}$
Explanation:
A Given that, Planck's constant, $(\mathrm{h})=6.62 \times 10^{34} \mathrm{Js}$ $\mathrm{V}=100 \text { Volt }$ Mass of electron, $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}$ $\lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$ $1=1.24 \times 10^{-10} \mathrm{~m}$ $1=0.124 \times 10^{-9} \mathrm{~m}$ $\lambda=0.12 \mathrm{~nm}$
142431
The wavelength ' $\lambda$ ' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is $(\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $h=$ Plank's constant)
1 $\frac{2 \lambda}{\mathrm{mch}}$
2 $\frac{\lambda \mathrm{mc}}{2 \mathrm{~h}}$
3 $\frac{\lambda \mathrm{m}}{4 \mathrm{~h}}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D The de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { So, } \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { Or } \quad \mathrm{v} =\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Energy of photon, $\mathrm{E}_{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}_{\mathrm{e}}=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{E}_{\mathrm{e}}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2} \mathrm{mv^{2 }}}=\frac{2 \mathrm{hc}}{\lambda \mathrm{mv}^{2}}$ Substituting the value of $\mathrm{v}$ in above equation, $\frac{E_{p}}{E_{e}}=\frac{2 h c}{\lambda m\left(\frac{h}{m \lambda}\right)^{2}}=\frac{2 \lambda m c}{h}$
MHT-CET 2020
Dual nature of radiation and Matter
142432
A photon and an electron have equal energy E. $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ is proportional to
1 $\sqrt{\mathrm{E}}$
2 $\frac{1}{\sqrt{\mathrm{E}}}$
3 $\frac{1}{\mathrm{E}}$
4 does not depend upon $\mathrm{E}$
Explanation:
B Wavelength of photon, $\lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$ Where, $\mathrm{h}=$ Planck constant $\mathrm{E}=$ Photon energy $\mathrm{c}=$ Speed of light $\lambda=$ Photon wavelength And wavelength of electron, $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Now, by equation (i) \(ii), we get- $\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} =\frac{\mathrm{hc}}{\mathrm{E}} \times \frac{\sqrt{2 \mathrm{mE}}}{\mathrm{h}}$ $=\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} \propto \frac{1}{\sqrt{\mathrm{E}}}$
Manipal UGET-2018
Dual nature of radiation and Matter
142433
An electron moving with initial velocity $\overrightarrow{\mathrm{V}}=\mathrm{V}_{0} \hat{\mathrm{i}}$ is moving in an magnetic field $\vec{B}=B_{0} \hat{J}$ then its de-Broglie wavelength
1 decreases with time
2 first increases and then decreases
3 increases with time
4 remains constant
Explanation:
D Given that, $\vec{V}=V_{0} \hat{i}$ $\vec{B}=B_{0} \hat{j}$ From de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{F} =\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=\mathrm{qVB} \hat{\mathrm{k}}$ The above force is perpendicular to $\vec{V}$ and $\vec{B}$ for the magnitude of $\mathrm{V}$ will not change. $\therefore$ Momentum (p) will not change So, $\lambda$ will also not change.
Karnataka CET-2019
Dual nature of radiation and Matter
142434
How much is the de-Broglie wavelength for an electron accelerated by an $100 \mathrm{~V}$ potential difference?
1 $0.123 \mathrm{~nm}$
2 $123 \mathrm{~nm}$
3 $12.3 \mathrm{~nm}$
4 $0.123 \mathrm{~cm}$
Explanation:
A Given that, Planck's constant, $(\mathrm{h})=6.62 \times 10^{34} \mathrm{Js}$ $\mathrm{V}=100 \text { Volt }$ Mass of electron, $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}$ $\lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$ $1=1.24 \times 10^{-10} \mathrm{~m}$ $1=0.124 \times 10^{-9} \mathrm{~m}$ $\lambda=0.12 \mathrm{~nm}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Dual nature of radiation and Matter
142431
The wavelength ' $\lambda$ ' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is $(\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $h=$ Plank's constant)
1 $\frac{2 \lambda}{\mathrm{mch}}$
2 $\frac{\lambda \mathrm{mc}}{2 \mathrm{~h}}$
3 $\frac{\lambda \mathrm{m}}{4 \mathrm{~h}}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D The de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { So, } \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { Or } \quad \mathrm{v} =\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Energy of photon, $\mathrm{E}_{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}_{\mathrm{e}}=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{E}_{\mathrm{e}}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2} \mathrm{mv^{2 }}}=\frac{2 \mathrm{hc}}{\lambda \mathrm{mv}^{2}}$ Substituting the value of $\mathrm{v}$ in above equation, $\frac{E_{p}}{E_{e}}=\frac{2 h c}{\lambda m\left(\frac{h}{m \lambda}\right)^{2}}=\frac{2 \lambda m c}{h}$
MHT-CET 2020
Dual nature of radiation and Matter
142432
A photon and an electron have equal energy E. $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ is proportional to
1 $\sqrt{\mathrm{E}}$
2 $\frac{1}{\sqrt{\mathrm{E}}}$
3 $\frac{1}{\mathrm{E}}$
4 does not depend upon $\mathrm{E}$
Explanation:
B Wavelength of photon, $\lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$ Where, $\mathrm{h}=$ Planck constant $\mathrm{E}=$ Photon energy $\mathrm{c}=$ Speed of light $\lambda=$ Photon wavelength And wavelength of electron, $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Now, by equation (i) \(ii), we get- $\frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} =\frac{\mathrm{hc}}{\mathrm{E}} \times \frac{\sqrt{2 \mathrm{mE}}}{\mathrm{h}}$ $=\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}$ $\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\mathrm{e}}} \propto \frac{1}{\sqrt{\mathrm{E}}}$
Manipal UGET-2018
Dual nature of radiation and Matter
142433
An electron moving with initial velocity $\overrightarrow{\mathrm{V}}=\mathrm{V}_{0} \hat{\mathrm{i}}$ is moving in an magnetic field $\vec{B}=B_{0} \hat{J}$ then its de-Broglie wavelength
1 decreases with time
2 first increases and then decreases
3 increases with time
4 remains constant
Explanation:
D Given that, $\vec{V}=V_{0} \hat{i}$ $\vec{B}=B_{0} \hat{j}$ From de-Broglie wavelength, $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{F} =\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=\mathrm{qVB} \hat{\mathrm{k}}$ The above force is perpendicular to $\vec{V}$ and $\vec{B}$ for the magnitude of $\mathrm{V}$ will not change. $\therefore$ Momentum (p) will not change So, $\lambda$ will also not change.
Karnataka CET-2019
Dual nature of radiation and Matter
142434
How much is the de-Broglie wavelength for an electron accelerated by an $100 \mathrm{~V}$ potential difference?
1 $0.123 \mathrm{~nm}$
2 $123 \mathrm{~nm}$
3 $12.3 \mathrm{~nm}$
4 $0.123 \mathrm{~cm}$
Explanation:
A Given that, Planck's constant, $(\mathrm{h})=6.62 \times 10^{34} \mathrm{Js}$ $\mathrm{V}=100 \text { Volt }$ Mass of electron, $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}$ $\lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$ $1=1.24 \times 10^{-10} \mathrm{~m}$ $1=0.124 \times 10^{-9} \mathrm{~m}$ $\lambda=0.12 \mathrm{~nm}$