142539
If an electron has an energy such that its deBroglie wavelength is $5500 \mathrm{~A}$, then the energy value of that electron is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right.$. $\mathrm{m}_{\mathrm{e}}=$ $9.1 \times 10^{-31} \mathrm{~kg}$ )
142540
If the velocity of a particle is reduced to half what is the percentage increase in its de-Broglie wavelength?
1 100
2 200
3 400
4 50
Explanation:
A We know that, de - Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Then, $\quad \lambda \propto \frac{1}{\mathrm{~V}}$ And, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{v}}{2}}$ $\lambda=\frac{2 \mathrm{~h}}{\mathrm{mv}}$ $\%$ Change $=\frac{\frac{2 \mathrm{~h}}{\mathrm{mv}}-\frac{\mathrm{h}}{\mathrm{mv}}}{\frac{\mathrm{h}}{\mathrm{mv}}} \times 100=100$
Dual nature of radiation and Matter
142541
The energy in $\mathrm{MeV}$ is released due to transformation of $1 \mathrm{~kg}$ mass completely into energy, is $\left(\mathrm{C}=\mathbf{3} \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
142542
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$.
142539
If an electron has an energy such that its deBroglie wavelength is $5500 \mathrm{~A}$, then the energy value of that electron is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right.$. $\mathrm{m}_{\mathrm{e}}=$ $9.1 \times 10^{-31} \mathrm{~kg}$ )
142540
If the velocity of a particle is reduced to half what is the percentage increase in its de-Broglie wavelength?
1 100
2 200
3 400
4 50
Explanation:
A We know that, de - Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Then, $\quad \lambda \propto \frac{1}{\mathrm{~V}}$ And, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{v}}{2}}$ $\lambda=\frac{2 \mathrm{~h}}{\mathrm{mv}}$ $\%$ Change $=\frac{\frac{2 \mathrm{~h}}{\mathrm{mv}}-\frac{\mathrm{h}}{\mathrm{mv}}}{\frac{\mathrm{h}}{\mathrm{mv}}} \times 100=100$
Dual nature of radiation and Matter
142541
The energy in $\mathrm{MeV}$ is released due to transformation of $1 \mathrm{~kg}$ mass completely into energy, is $\left(\mathrm{C}=\mathbf{3} \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
142542
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$.
142539
If an electron has an energy such that its deBroglie wavelength is $5500 \mathrm{~A}$, then the energy value of that electron is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right.$. $\mathrm{m}_{\mathrm{e}}=$ $9.1 \times 10^{-31} \mathrm{~kg}$ )
142540
If the velocity of a particle is reduced to half what is the percentage increase in its de-Broglie wavelength?
1 100
2 200
3 400
4 50
Explanation:
A We know that, de - Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Then, $\quad \lambda \propto \frac{1}{\mathrm{~V}}$ And, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{v}}{2}}$ $\lambda=\frac{2 \mathrm{~h}}{\mathrm{mv}}$ $\%$ Change $=\frac{\frac{2 \mathrm{~h}}{\mathrm{mv}}-\frac{\mathrm{h}}{\mathrm{mv}}}{\frac{\mathrm{h}}{\mathrm{mv}}} \times 100=100$
Dual nature of radiation and Matter
142541
The energy in $\mathrm{MeV}$ is released due to transformation of $1 \mathrm{~kg}$ mass completely into energy, is $\left(\mathrm{C}=\mathbf{3} \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
142542
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$.
142539
If an electron has an energy such that its deBroglie wavelength is $5500 \mathrm{~A}$, then the energy value of that electron is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right.$. $\mathrm{m}_{\mathrm{e}}=$ $9.1 \times 10^{-31} \mathrm{~kg}$ )
142540
If the velocity of a particle is reduced to half what is the percentage increase in its de-Broglie wavelength?
1 100
2 200
3 400
4 50
Explanation:
A We know that, de - Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Then, $\quad \lambda \propto \frac{1}{\mathrm{~V}}$ And, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{v}}{2}}$ $\lambda=\frac{2 \mathrm{~h}}{\mathrm{mv}}$ $\%$ Change $=\frac{\frac{2 \mathrm{~h}}{\mathrm{mv}}-\frac{\mathrm{h}}{\mathrm{mv}}}{\frac{\mathrm{h}}{\mathrm{mv}}} \times 100=100$
Dual nature of radiation and Matter
142541
The energy in $\mathrm{MeV}$ is released due to transformation of $1 \mathrm{~kg}$ mass completely into energy, is $\left(\mathrm{C}=\mathbf{3} \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
142542
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$.
142539
If an electron has an energy such that its deBroglie wavelength is $5500 \mathrm{~A}$, then the energy value of that electron is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right.$. $\mathrm{m}_{\mathrm{e}}=$ $9.1 \times 10^{-31} \mathrm{~kg}$ )
142540
If the velocity of a particle is reduced to half what is the percentage increase in its de-Broglie wavelength?
1 100
2 200
3 400
4 50
Explanation:
A We know that, de - Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Then, $\quad \lambda \propto \frac{1}{\mathrm{~V}}$ And, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{v}}{2}}$ $\lambda=\frac{2 \mathrm{~h}}{\mathrm{mv}}$ $\%$ Change $=\frac{\frac{2 \mathrm{~h}}{\mathrm{mv}}-\frac{\mathrm{h}}{\mathrm{mv}}}{\frac{\mathrm{h}}{\mathrm{mv}}} \times 100=100$
Dual nature of radiation and Matter
142541
The energy in $\mathrm{MeV}$ is released due to transformation of $1 \mathrm{~kg}$ mass completely into energy, is $\left(\mathrm{C}=\mathbf{3} \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
142542
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$.
