238833
Given : The mass of electron is $9.11 \times 10^{-31} \mathrm{~kg}$. Planck constant is $6.626 \times 10^{-34} \mathrm{Js}$, the uncertainty involved in the measurement of velocity within a distance of $0.1 \AA \dot{\AA}$ is
1 $5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}$
2 $5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$
3 $5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
4 $5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Explanation:
According to Heisenberg's uncertainty principle, $\Delta \mathrm{x} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{x}}$ Where $\Delta \mathrm{x}$ is uncertainty involved in the measurement of position $\Delta \mathrm{v}$ is uncertainty involved in the measurement of velocity $h$ is planck's constant $m$ is mass of the electron $\begin{aligned} & \because \quad \text { Given } \Delta \mathrm{x}=0.1 \AA \\ &=\left(0.1 \times 10^{-10}\right) \\ &=10^{-11} \mathrm{~m} \\ & \mathrm{~m}=9.11 \times 10^{-31} \mathrm{~kg} \\ & \mathrm{~h}=6.626 \times 10^{-34} \mathrm{Js} \end{aligned}$ On substituting the values- $\begin{aligned} & \Delta \mathrm{v}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}} \\ & =0.0579 \times 10^8 \mathrm{~ms}^{-1}=5.79 \times 10^6 \mathrm{~ms}^{-1} \end{aligned}$
NEET-2006
Structure of Atom
238834
The uncertainty in momentum of an electron is $1 \times 10^{-5} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. The uncertainty in its position will be $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\right)$
1 $5.27 \times 10^{-30} \mathrm{~m}$
2 $1.05 \times 10^{-26} \mathrm{~m}$
3 $1.05 \times 10^{-28} \mathrm{~m}$
4 $5.25 \times 10^{-28} \mathrm{~m}$
Explanation:
The uncertainty in the position of an electron is : $\Delta x=\frac{h}{4 \pi \Delta \mathrm{p}}$ Where, $\Delta \mathrm{x}=$ The position of the particle $\Delta \mathrm{p}=$ The momentum of the particle $\mathrm{h}=$ Planck's constant $\begin{aligned} \Delta x & =\frac{6.626 \times 10^{-34}}{4 \times 3.146 \times 10^{-5}} \\ & =5.27 \times 10^{-30} \mathrm{~m} \end{aligned}$
NEET-1998
Structure of Atom
238836
Uncertainty in position of an electron (mass = $9.1 \times 10^{-28}$ g) moving with a velocity of $3 \times 10^4$ $\mathrm{cm} / \mathrm{s}$ accurate upto $0.001 \%$ will be (use $h /(4 \pi)$ in uncertainity expression where $h$ $=6.626 \times 10^{-27}$ erg second)
1 $5.76 \mathrm{~cm}$
2 $7.68 \mathrm{~cm}$
3 $1.93 \mathrm{~cm}$
4 $3.84 \mathrm{~cm}$
Explanation:
According to Heisenberg's uncertainty principle $\Delta \mathrm{x} \times \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ Where, \(\Delta \mathrm{x}=\) Uncertainty in position or change in position \(\Delta \mathrm{v}=\) uncertainty in velocity \(\mathrm{h}=\) Planck' \(\mathrm{s}\) constant \(\left(6.626 \times 10^{-27} \mathrm{Js}\right)\) \(\mathrm{m}=\) mass of electron \(\left(9.1 \times 10^{-28} \mathrm{~kg}\right)\) Here, \(\Delta \mathrm{v}=0.001 \%\) of \(3 \times 10^4\) \(=\frac{0.001}{100} \times 3 \times 10^4=0.3 \mathrm{~cm} / \mathrm{s}\) \(\therefore \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}}\) \(=\frac{6.626 \times 10^{-27}}{4 \times 3.14 \times 9.1 \times 10^{-28} \times 0.3}=1.93 \mathrm{~cm}\)
NEET-1995
Structure of Atom
238837
If $\mathbf{E}_e, \mathbf{E}_\alpha$ and $\mathbf{E}_{\mathrm{p}}$ represent the kinetic energies of an electron, $\alpha$-particle and a proton respectively each moving with same de-Broglie wavelength then
238833
Given : The mass of electron is $9.11 \times 10^{-31} \mathrm{~kg}$. Planck constant is $6.626 \times 10^{-34} \mathrm{Js}$, the uncertainty involved in the measurement of velocity within a distance of $0.1 \AA \dot{\AA}$ is
1 $5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}$
2 $5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$
3 $5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
4 $5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Explanation:
According to Heisenberg's uncertainty principle, $\Delta \mathrm{x} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{x}}$ Where $\Delta \mathrm{x}$ is uncertainty involved in the measurement of position $\Delta \mathrm{v}$ is uncertainty involved in the measurement of velocity $h$ is planck's constant $m$ is mass of the electron $\begin{aligned} & \because \quad \text { Given } \Delta \mathrm{x}=0.1 \AA \\ &=\left(0.1 \times 10^{-10}\right) \\ &=10^{-11} \mathrm{~m} \\ & \mathrm{~m}=9.11 \times 10^{-31} \mathrm{~kg} \\ & \mathrm{~h}=6.626 \times 10^{-34} \mathrm{Js} \end{aligned}$ On substituting the values- $\begin{aligned} & \Delta \mathrm{v}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}} \\ & =0.0579 \times 10^8 \mathrm{~ms}^{-1}=5.79 \times 10^6 \mathrm{~ms}^{-1} \end{aligned}$
NEET-2006
Structure of Atom
238834
The uncertainty in momentum of an electron is $1 \times 10^{-5} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. The uncertainty in its position will be $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\right)$
1 $5.27 \times 10^{-30} \mathrm{~m}$
2 $1.05 \times 10^{-26} \mathrm{~m}$
3 $1.05 \times 10^{-28} \mathrm{~m}$
4 $5.25 \times 10^{-28} \mathrm{~m}$
Explanation:
The uncertainty in the position of an electron is : $\Delta x=\frac{h}{4 \pi \Delta \mathrm{p}}$ Where, $\Delta \mathrm{x}=$ The position of the particle $\Delta \mathrm{p}=$ The momentum of the particle $\mathrm{h}=$ Planck's constant $\begin{aligned} \Delta x & =\frac{6.626 \times 10^{-34}}{4 \times 3.146 \times 10^{-5}} \\ & =5.27 \times 10^{-30} \mathrm{~m} \end{aligned}$
NEET-1998
Structure of Atom
238836
Uncertainty in position of an electron (mass = $9.1 \times 10^{-28}$ g) moving with a velocity of $3 \times 10^4$ $\mathrm{cm} / \mathrm{s}$ accurate upto $0.001 \%$ will be (use $h /(4 \pi)$ in uncertainity expression where $h$ $=6.626 \times 10^{-27}$ erg second)
1 $5.76 \mathrm{~cm}$
2 $7.68 \mathrm{~cm}$
3 $1.93 \mathrm{~cm}$
4 $3.84 \mathrm{~cm}$
Explanation:
According to Heisenberg's uncertainty principle $\Delta \mathrm{x} \times \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ Where, \(\Delta \mathrm{x}=\) Uncertainty in position or change in position \(\Delta \mathrm{v}=\) uncertainty in velocity \(\mathrm{h}=\) Planck' \(\mathrm{s}\) constant \(\left(6.626 \times 10^{-27} \mathrm{Js}\right)\) \(\mathrm{m}=\) mass of electron \(\left(9.1 \times 10^{-28} \mathrm{~kg}\right)\) Here, \(\Delta \mathrm{v}=0.001 \%\) of \(3 \times 10^4\) \(=\frac{0.001}{100} \times 3 \times 10^4=0.3 \mathrm{~cm} / \mathrm{s}\) \(\therefore \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}}\) \(=\frac{6.626 \times 10^{-27}}{4 \times 3.14 \times 9.1 \times 10^{-28} \times 0.3}=1.93 \mathrm{~cm}\)
NEET-1995
Structure of Atom
238837
If $\mathbf{E}_e, \mathbf{E}_\alpha$ and $\mathbf{E}_{\mathrm{p}}$ represent the kinetic energies of an electron, $\alpha$-particle and a proton respectively each moving with same de-Broglie wavelength then
238833
Given : The mass of electron is $9.11 \times 10^{-31} \mathrm{~kg}$. Planck constant is $6.626 \times 10^{-34} \mathrm{Js}$, the uncertainty involved in the measurement of velocity within a distance of $0.1 \AA \dot{\AA}$ is
1 $5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}$
2 $5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$
3 $5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
4 $5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Explanation:
According to Heisenberg's uncertainty principle, $\Delta \mathrm{x} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{x}}$ Where $\Delta \mathrm{x}$ is uncertainty involved in the measurement of position $\Delta \mathrm{v}$ is uncertainty involved in the measurement of velocity $h$ is planck's constant $m$ is mass of the electron $\begin{aligned} & \because \quad \text { Given } \Delta \mathrm{x}=0.1 \AA \\ &=\left(0.1 \times 10^{-10}\right) \\ &=10^{-11} \mathrm{~m} \\ & \mathrm{~m}=9.11 \times 10^{-31} \mathrm{~kg} \\ & \mathrm{~h}=6.626 \times 10^{-34} \mathrm{Js} \end{aligned}$ On substituting the values- $\begin{aligned} & \Delta \mathrm{v}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}} \\ & =0.0579 \times 10^8 \mathrm{~ms}^{-1}=5.79 \times 10^6 \mathrm{~ms}^{-1} \end{aligned}$
NEET-2006
Structure of Atom
238834
The uncertainty in momentum of an electron is $1 \times 10^{-5} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. The uncertainty in its position will be $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\right)$
1 $5.27 \times 10^{-30} \mathrm{~m}$
2 $1.05 \times 10^{-26} \mathrm{~m}$
3 $1.05 \times 10^{-28} \mathrm{~m}$
4 $5.25 \times 10^{-28} \mathrm{~m}$
Explanation:
The uncertainty in the position of an electron is : $\Delta x=\frac{h}{4 \pi \Delta \mathrm{p}}$ Where, $\Delta \mathrm{x}=$ The position of the particle $\Delta \mathrm{p}=$ The momentum of the particle $\mathrm{h}=$ Planck's constant $\begin{aligned} \Delta x & =\frac{6.626 \times 10^{-34}}{4 \times 3.146 \times 10^{-5}} \\ & =5.27 \times 10^{-30} \mathrm{~m} \end{aligned}$
NEET-1998
Structure of Atom
238836
Uncertainty in position of an electron (mass = $9.1 \times 10^{-28}$ g) moving with a velocity of $3 \times 10^4$ $\mathrm{cm} / \mathrm{s}$ accurate upto $0.001 \%$ will be (use $h /(4 \pi)$ in uncertainity expression where $h$ $=6.626 \times 10^{-27}$ erg second)
1 $5.76 \mathrm{~cm}$
2 $7.68 \mathrm{~cm}$
3 $1.93 \mathrm{~cm}$
4 $3.84 \mathrm{~cm}$
Explanation:
According to Heisenberg's uncertainty principle $\Delta \mathrm{x} \times \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ Where, \(\Delta \mathrm{x}=\) Uncertainty in position or change in position \(\Delta \mathrm{v}=\) uncertainty in velocity \(\mathrm{h}=\) Planck' \(\mathrm{s}\) constant \(\left(6.626 \times 10^{-27} \mathrm{Js}\right)\) \(\mathrm{m}=\) mass of electron \(\left(9.1 \times 10^{-28} \mathrm{~kg}\right)\) Here, \(\Delta \mathrm{v}=0.001 \%\) of \(3 \times 10^4\) \(=\frac{0.001}{100} \times 3 \times 10^4=0.3 \mathrm{~cm} / \mathrm{s}\) \(\therefore \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}}\) \(=\frac{6.626 \times 10^{-27}}{4 \times 3.14 \times 9.1 \times 10^{-28} \times 0.3}=1.93 \mathrm{~cm}\)
NEET-1995
Structure of Atom
238837
If $\mathbf{E}_e, \mathbf{E}_\alpha$ and $\mathbf{E}_{\mathrm{p}}$ represent the kinetic energies of an electron, $\alpha$-particle and a proton respectively each moving with same de-Broglie wavelength then
238833
Given : The mass of electron is $9.11 \times 10^{-31} \mathrm{~kg}$. Planck constant is $6.626 \times 10^{-34} \mathrm{Js}$, the uncertainty involved in the measurement of velocity within a distance of $0.1 \AA \dot{\AA}$ is
1 $5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}$
2 $5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$
3 $5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
4 $5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Explanation:
According to Heisenberg's uncertainty principle, $\Delta \mathrm{x} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{x}}$ Where $\Delta \mathrm{x}$ is uncertainty involved in the measurement of position $\Delta \mathrm{v}$ is uncertainty involved in the measurement of velocity $h$ is planck's constant $m$ is mass of the electron $\begin{aligned} & \because \quad \text { Given } \Delta \mathrm{x}=0.1 \AA \\ &=\left(0.1 \times 10^{-10}\right) \\ &=10^{-11} \mathrm{~m} \\ & \mathrm{~m}=9.11 \times 10^{-31} \mathrm{~kg} \\ & \mathrm{~h}=6.626 \times 10^{-34} \mathrm{Js} \end{aligned}$ On substituting the values- $\begin{aligned} & \Delta \mathrm{v}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}} \\ & =0.0579 \times 10^8 \mathrm{~ms}^{-1}=5.79 \times 10^6 \mathrm{~ms}^{-1} \end{aligned}$
NEET-2006
Structure of Atom
238834
The uncertainty in momentum of an electron is $1 \times 10^{-5} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. The uncertainty in its position will be $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\right)$
1 $5.27 \times 10^{-30} \mathrm{~m}$
2 $1.05 \times 10^{-26} \mathrm{~m}$
3 $1.05 \times 10^{-28} \mathrm{~m}$
4 $5.25 \times 10^{-28} \mathrm{~m}$
Explanation:
The uncertainty in the position of an electron is : $\Delta x=\frac{h}{4 \pi \Delta \mathrm{p}}$ Where, $\Delta \mathrm{x}=$ The position of the particle $\Delta \mathrm{p}=$ The momentum of the particle $\mathrm{h}=$ Planck's constant $\begin{aligned} \Delta x & =\frac{6.626 \times 10^{-34}}{4 \times 3.146 \times 10^{-5}} \\ & =5.27 \times 10^{-30} \mathrm{~m} \end{aligned}$
NEET-1998
Structure of Atom
238836
Uncertainty in position of an electron (mass = $9.1 \times 10^{-28}$ g) moving with a velocity of $3 \times 10^4$ $\mathrm{cm} / \mathrm{s}$ accurate upto $0.001 \%$ will be (use $h /(4 \pi)$ in uncertainity expression where $h$ $=6.626 \times 10^{-27}$ erg second)
1 $5.76 \mathrm{~cm}$
2 $7.68 \mathrm{~cm}$
3 $1.93 \mathrm{~cm}$
4 $3.84 \mathrm{~cm}$
Explanation:
According to Heisenberg's uncertainty principle $\Delta \mathrm{x} \times \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ Where, \(\Delta \mathrm{x}=\) Uncertainty in position or change in position \(\Delta \mathrm{v}=\) uncertainty in velocity \(\mathrm{h}=\) Planck' \(\mathrm{s}\) constant \(\left(6.626 \times 10^{-27} \mathrm{Js}\right)\) \(\mathrm{m}=\) mass of electron \(\left(9.1 \times 10^{-28} \mathrm{~kg}\right)\) Here, \(\Delta \mathrm{v}=0.001 \%\) of \(3 \times 10^4\) \(=\frac{0.001}{100} \times 3 \times 10^4=0.3 \mathrm{~cm} / \mathrm{s}\) \(\therefore \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}}\) \(=\frac{6.626 \times 10^{-27}}{4 \times 3.14 \times 9.1 \times 10^{-28} \times 0.3}=1.93 \mathrm{~cm}\)
NEET-1995
Structure of Atom
238837
If $\mathbf{E}_e, \mathbf{E}_\alpha$ and $\mathbf{E}_{\mathrm{p}}$ represent the kinetic energies of an electron, $\alpha$-particle and a proton respectively each moving with same de-Broglie wavelength then
