238798
A fast moving particle of mass $6.63 \times 10^{-28}$ can be located with an accuracy of $1 \AA$. The uncertainty in its velocity (in $\left.\mathrm{ms}^{-1}\right)$ is about $(h=$ $\left.6.63 \times 10^{-34} \mathrm{Js}\right)$
238799
The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 \mathrm{~ms}^{-1}$ respectively. The mass of $B$ is five times to that of $A$. What is the ratio of uncertainties $\left(\frac{\Delta x_A}{\Delta x_B}\right)$ their positions in :
1 2
2 0.25
3 4
4 1
Explanation:
Given, $\Delta \mathrm{v}_{\mathrm{A}}=0.05 \mathrm{~ms}^{-1}$ and $\mathrm{m}_{\mathrm{A}}=\mathrm{m}$ $\Delta \mathrm{v}_{\mathrm{B}}=0.02 \mathrm{~ms}^{-1} \quad \mathrm{~m}_{\mathrm{B}}=5 \mathrm{~m}$ According to Heisenberg uncertainty principle- $\mathrm{m} \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$ Where, $\mathrm{h}=$ Planck's constant $\begin{aligned} & \Delta \mathrm{x}=\text { Uncertainity in position } \\ & \Delta \mathrm{v}=\text { Uncertainity in velocity } \\ & \mathrm{m}=\text { Mass } \end{aligned}$ For particle A : $\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}=\frac{\mathrm{h}}{4 \pi}$ For particle B : $5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}=\frac{\mathrm{h}}{4 \pi}$ Dividing equation (i) and (ii) we get : $\begin{gathered} \frac{\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}}{5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}}=1 \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=\frac{0.02 \times 5}{0.05} \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=2 \end{gathered}$
AP-EAMCET-2006
Structure of Atom
238800
If the uncertainty in velocity of a moving object is $1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ and the uncertainty in its position is $58 \mathrm{~m}$, The mass of this object is approximately equal to that of $\left(h=6.626 \times 10^{-34} \mathrm{Js}\right)$
1 helium
2 deuterium
3 lithium
4 electron
Explanation:
Given- Uncertainty in velocity $(\Delta \mathrm{v})=1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ Uncertainty in position $(\Delta \mathrm{x})=58 \mathrm{~m}$ $\text { Mass }(\mathrm{m})=\text { ? }$ Now, from the Heisenberg principle- $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$ where $-\Delta x=$ uncertainty in position $\Delta \mathrm{p}=$ uncertainty in momentum $\therefore \Delta \mathrm{x} . \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\mathrm{m}=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{-6}}$ or or Thus, the mass of moving object is approximately equal to that of electron.
AP EAMCET (Medical) - 2013
Structure of Atom
238801
If a cricket ball of weight $0.1 \mathrm{~kg}$ has an uncertainty of $0.15 \mathrm{~ms}^{-1}$ in its velocity then the uncertainty in the position of the cricket balls is
238803
Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^4 \mathrm{~cm} . \mathrm{s}^{-1}$. accurate up to $0.0016 \%$. (Given, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-28}$ g. $\mathrm{h}=6.626 \times 10^{-27}$ erg.s)
238798
A fast moving particle of mass $6.63 \times 10^{-28}$ can be located with an accuracy of $1 \AA$. The uncertainty in its velocity (in $\left.\mathrm{ms}^{-1}\right)$ is about $(h=$ $\left.6.63 \times 10^{-34} \mathrm{Js}\right)$
238799
The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 \mathrm{~ms}^{-1}$ respectively. The mass of $B$ is five times to that of $A$. What is the ratio of uncertainties $\left(\frac{\Delta x_A}{\Delta x_B}\right)$ their positions in :
1 2
2 0.25
3 4
4 1
Explanation:
Given, $\Delta \mathrm{v}_{\mathrm{A}}=0.05 \mathrm{~ms}^{-1}$ and $\mathrm{m}_{\mathrm{A}}=\mathrm{m}$ $\Delta \mathrm{v}_{\mathrm{B}}=0.02 \mathrm{~ms}^{-1} \quad \mathrm{~m}_{\mathrm{B}}=5 \mathrm{~m}$ According to Heisenberg uncertainty principle- $\mathrm{m} \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$ Where, $\mathrm{h}=$ Planck's constant $\begin{aligned} & \Delta \mathrm{x}=\text { Uncertainity in position } \\ & \Delta \mathrm{v}=\text { Uncertainity in velocity } \\ & \mathrm{m}=\text { Mass } \end{aligned}$ For particle A : $\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}=\frac{\mathrm{h}}{4 \pi}$ For particle B : $5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}=\frac{\mathrm{h}}{4 \pi}$ Dividing equation (i) and (ii) we get : $\begin{gathered} \frac{\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}}{5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}}=1 \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=\frac{0.02 \times 5}{0.05} \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=2 \end{gathered}$
AP-EAMCET-2006
Structure of Atom
238800
If the uncertainty in velocity of a moving object is $1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ and the uncertainty in its position is $58 \mathrm{~m}$, The mass of this object is approximately equal to that of $\left(h=6.626 \times 10^{-34} \mathrm{Js}\right)$
1 helium
2 deuterium
3 lithium
4 electron
Explanation:
Given- Uncertainty in velocity $(\Delta \mathrm{v})=1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ Uncertainty in position $(\Delta \mathrm{x})=58 \mathrm{~m}$ $\text { Mass }(\mathrm{m})=\text { ? }$ Now, from the Heisenberg principle- $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$ where $-\Delta x=$ uncertainty in position $\Delta \mathrm{p}=$ uncertainty in momentum $\therefore \Delta \mathrm{x} . \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\mathrm{m}=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{-6}}$ or or Thus, the mass of moving object is approximately equal to that of electron.
AP EAMCET (Medical) - 2013
Structure of Atom
238801
If a cricket ball of weight $0.1 \mathrm{~kg}$ has an uncertainty of $0.15 \mathrm{~ms}^{-1}$ in its velocity then the uncertainty in the position of the cricket balls is
238803
Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^4 \mathrm{~cm} . \mathrm{s}^{-1}$. accurate up to $0.0016 \%$. (Given, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-28}$ g. $\mathrm{h}=6.626 \times 10^{-27}$ erg.s)
238798
A fast moving particle of mass $6.63 \times 10^{-28}$ can be located with an accuracy of $1 \AA$. The uncertainty in its velocity (in $\left.\mathrm{ms}^{-1}\right)$ is about $(h=$ $\left.6.63 \times 10^{-34} \mathrm{Js}\right)$
238799
The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 \mathrm{~ms}^{-1}$ respectively. The mass of $B$ is five times to that of $A$. What is the ratio of uncertainties $\left(\frac{\Delta x_A}{\Delta x_B}\right)$ their positions in :
1 2
2 0.25
3 4
4 1
Explanation:
Given, $\Delta \mathrm{v}_{\mathrm{A}}=0.05 \mathrm{~ms}^{-1}$ and $\mathrm{m}_{\mathrm{A}}=\mathrm{m}$ $\Delta \mathrm{v}_{\mathrm{B}}=0.02 \mathrm{~ms}^{-1} \quad \mathrm{~m}_{\mathrm{B}}=5 \mathrm{~m}$ According to Heisenberg uncertainty principle- $\mathrm{m} \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$ Where, $\mathrm{h}=$ Planck's constant $\begin{aligned} & \Delta \mathrm{x}=\text { Uncertainity in position } \\ & \Delta \mathrm{v}=\text { Uncertainity in velocity } \\ & \mathrm{m}=\text { Mass } \end{aligned}$ For particle A : $\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}=\frac{\mathrm{h}}{4 \pi}$ For particle B : $5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}=\frac{\mathrm{h}}{4 \pi}$ Dividing equation (i) and (ii) we get : $\begin{gathered} \frac{\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}}{5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}}=1 \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=\frac{0.02 \times 5}{0.05} \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=2 \end{gathered}$
AP-EAMCET-2006
Structure of Atom
238800
If the uncertainty in velocity of a moving object is $1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ and the uncertainty in its position is $58 \mathrm{~m}$, The mass of this object is approximately equal to that of $\left(h=6.626 \times 10^{-34} \mathrm{Js}\right)$
1 helium
2 deuterium
3 lithium
4 electron
Explanation:
Given- Uncertainty in velocity $(\Delta \mathrm{v})=1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ Uncertainty in position $(\Delta \mathrm{x})=58 \mathrm{~m}$ $\text { Mass }(\mathrm{m})=\text { ? }$ Now, from the Heisenberg principle- $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$ where $-\Delta x=$ uncertainty in position $\Delta \mathrm{p}=$ uncertainty in momentum $\therefore \Delta \mathrm{x} . \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\mathrm{m}=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{-6}}$ or or Thus, the mass of moving object is approximately equal to that of electron.
AP EAMCET (Medical) - 2013
Structure of Atom
238801
If a cricket ball of weight $0.1 \mathrm{~kg}$ has an uncertainty of $0.15 \mathrm{~ms}^{-1}$ in its velocity then the uncertainty in the position of the cricket balls is
238803
Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^4 \mathrm{~cm} . \mathrm{s}^{-1}$. accurate up to $0.0016 \%$. (Given, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-28}$ g. $\mathrm{h}=6.626 \times 10^{-27}$ erg.s)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Structure of Atom
238798
A fast moving particle of mass $6.63 \times 10^{-28}$ can be located with an accuracy of $1 \AA$. The uncertainty in its velocity (in $\left.\mathrm{ms}^{-1}\right)$ is about $(h=$ $\left.6.63 \times 10^{-34} \mathrm{Js}\right)$
238799
The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 \mathrm{~ms}^{-1}$ respectively. The mass of $B$ is five times to that of $A$. What is the ratio of uncertainties $\left(\frac{\Delta x_A}{\Delta x_B}\right)$ their positions in :
1 2
2 0.25
3 4
4 1
Explanation:
Given, $\Delta \mathrm{v}_{\mathrm{A}}=0.05 \mathrm{~ms}^{-1}$ and $\mathrm{m}_{\mathrm{A}}=\mathrm{m}$ $\Delta \mathrm{v}_{\mathrm{B}}=0.02 \mathrm{~ms}^{-1} \quad \mathrm{~m}_{\mathrm{B}}=5 \mathrm{~m}$ According to Heisenberg uncertainty principle- $\mathrm{m} \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$ Where, $\mathrm{h}=$ Planck's constant $\begin{aligned} & \Delta \mathrm{x}=\text { Uncertainity in position } \\ & \Delta \mathrm{v}=\text { Uncertainity in velocity } \\ & \mathrm{m}=\text { Mass } \end{aligned}$ For particle A : $\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}=\frac{\mathrm{h}}{4 \pi}$ For particle B : $5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}=\frac{\mathrm{h}}{4 \pi}$ Dividing equation (i) and (ii) we get : $\begin{gathered} \frac{\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}}{5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}}=1 \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=\frac{0.02 \times 5}{0.05} \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=2 \end{gathered}$
AP-EAMCET-2006
Structure of Atom
238800
If the uncertainty in velocity of a moving object is $1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ and the uncertainty in its position is $58 \mathrm{~m}$, The mass of this object is approximately equal to that of $\left(h=6.626 \times 10^{-34} \mathrm{Js}\right)$
1 helium
2 deuterium
3 lithium
4 electron
Explanation:
Given- Uncertainty in velocity $(\Delta \mathrm{v})=1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ Uncertainty in position $(\Delta \mathrm{x})=58 \mathrm{~m}$ $\text { Mass }(\mathrm{m})=\text { ? }$ Now, from the Heisenberg principle- $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$ where $-\Delta x=$ uncertainty in position $\Delta \mathrm{p}=$ uncertainty in momentum $\therefore \Delta \mathrm{x} . \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\mathrm{m}=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{-6}}$ or or Thus, the mass of moving object is approximately equal to that of electron.
AP EAMCET (Medical) - 2013
Structure of Atom
238801
If a cricket ball of weight $0.1 \mathrm{~kg}$ has an uncertainty of $0.15 \mathrm{~ms}^{-1}$ in its velocity then the uncertainty in the position of the cricket balls is
238803
Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^4 \mathrm{~cm} . \mathrm{s}^{-1}$. accurate up to $0.0016 \%$. (Given, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-28}$ g. $\mathrm{h}=6.626 \times 10^{-27}$ erg.s)
238798
A fast moving particle of mass $6.63 \times 10^{-28}$ can be located with an accuracy of $1 \AA$. The uncertainty in its velocity (in $\left.\mathrm{ms}^{-1}\right)$ is about $(h=$ $\left.6.63 \times 10^{-34} \mathrm{Js}\right)$
238799
The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 \mathrm{~ms}^{-1}$ respectively. The mass of $B$ is five times to that of $A$. What is the ratio of uncertainties $\left(\frac{\Delta x_A}{\Delta x_B}\right)$ their positions in :
1 2
2 0.25
3 4
4 1
Explanation:
Given, $\Delta \mathrm{v}_{\mathrm{A}}=0.05 \mathrm{~ms}^{-1}$ and $\mathrm{m}_{\mathrm{A}}=\mathrm{m}$ $\Delta \mathrm{v}_{\mathrm{B}}=0.02 \mathrm{~ms}^{-1} \quad \mathrm{~m}_{\mathrm{B}}=5 \mathrm{~m}$ According to Heisenberg uncertainty principle- $\mathrm{m} \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$ Where, $\mathrm{h}=$ Planck's constant $\begin{aligned} & \Delta \mathrm{x}=\text { Uncertainity in position } \\ & \Delta \mathrm{v}=\text { Uncertainity in velocity } \\ & \mathrm{m}=\text { Mass } \end{aligned}$ For particle A : $\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}=\frac{\mathrm{h}}{4 \pi}$ For particle B : $5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}=\frac{\mathrm{h}}{4 \pi}$ Dividing equation (i) and (ii) we get : $\begin{gathered} \frac{\mathrm{m} \times 0.05 \times \Delta \mathrm{x}_{\mathrm{A}}}{5 \mathrm{~m} \times 0.02 \times \Delta \mathrm{x}_{\mathrm{B}}}=1 \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=\frac{0.02 \times 5}{0.05} \\ \frac{\Delta \mathrm{x}_{\mathrm{A}}}{\Delta \mathrm{x}_{\mathrm{B}}}=2 \end{gathered}$
AP-EAMCET-2006
Structure of Atom
238800
If the uncertainty in velocity of a moving object is $1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ and the uncertainty in its position is $58 \mathrm{~m}$, The mass of this object is approximately equal to that of $\left(h=6.626 \times 10^{-34} \mathrm{Js}\right)$
1 helium
2 deuterium
3 lithium
4 electron
Explanation:
Given- Uncertainty in velocity $(\Delta \mathrm{v})=1.0 \times 10^{-6} \mathrm{~ms}^{-1}$ Uncertainty in position $(\Delta \mathrm{x})=58 \mathrm{~m}$ $\text { Mass }(\mathrm{m})=\text { ? }$ Now, from the Heisenberg principle- $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$ where $-\Delta x=$ uncertainty in position $\Delta \mathrm{p}=$ uncertainty in momentum $\therefore \Delta \mathrm{x} . \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}$ or $\mathrm{m}=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{-6}}$ or or Thus, the mass of moving object is approximately equal to that of electron.
AP EAMCET (Medical) - 2013
Structure of Atom
238801
If a cricket ball of weight $0.1 \mathrm{~kg}$ has an uncertainty of $0.15 \mathrm{~ms}^{-1}$ in its velocity then the uncertainty in the position of the cricket balls is
238803
Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^4 \mathrm{~cm} . \mathrm{s}^{-1}$. accurate up to $0.0016 \%$. (Given, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-28}$ g. $\mathrm{h}=6.626 \times 10^{-27}$ erg.s)