QBManager

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 ${ms}^{- 1})$ is about $(h =$ $6.63 \times 10^{- 34} Js)$

1

8 \times 10^{3}

2

8 \times 10^{4}

3

8 \times 10^{5}

4

8 \times 10^{6}

5

8 \times 10^{7}

Explanation:

$\begin{aligned} G i v e n t h a t - \\ M a s s o f p a r t i c l e $ (m) = 6.63 \times 10^{- 28} g \\ = 6.63 \times 10^{- 31} kg \\ P l a n c k^{'} s c o n s t a n t (h) = 6.63 \times 10^{- 34} Js \\ Δ x = 1 A^{\circ} = 10^{- 10} m \end{aligned}$
According to Heisenberg's uncertainty principle
$\begin{aligned} Δ x . Δ p = \frac{h}{4 π} \\ Δ x . m Δ v = \frac{h}{4 π} \\ Δ v = \frac{h}{4 π \times m \times Δ x} \\ Δ v = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 6.63 \times 10^{- 31} \times 10^{- 10}} \\ Δ v = \frac{10^{- 34} \times 10^{41}}{12.56} \\ Δ v = \frac{10^{7}}{12.56} \\ Δ v = \frac{100 \times 10^{5}}{12.56} \\ Δ v = 8 \times 10^{5} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238799 The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 {ms}^{- 1}$ respectively. The mass of $B$ is five times to that of $A$ . What is the ratio of uncertainties $(\frac{Δ x_{A}}{Δ x_{B}})$ their positions in :

1 2

2 0.25

3 4

4 1

Explanation:

Given, $Δ v_{A} = 0.05 {ms}^{- 1}$ and $m_{A} = m$
$Δ v_{B} = 0.02 {ms}^{- 1} m_{B} = 5 m$
According to Heisenberg uncertainty principle-
$m Δ v \cdot Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant
$\begin{aligned} Δ x = Uncertainity in position \\ Δ v = Uncertainity in velocity \\ m = Mass \end{aligned}$
For particle A :
$m \times 0.05 \times Δ x_{A} = \frac{h}{4 π}$
For particle B :
$5 m \times 0.02 \times Δ x_{B} = \frac{h}{4 π}$
Dividing equation (i) and (ii) we get :
$\begin{matrix} \frac{m \times 0.05 \times Δ x_{A}}{5 m \times 0.02 \times Δ x_{B}} = 1 \\ \frac{Δ x_{A}}{Δ x_{B}} = \frac{0.02 \times 5}{0.05} \\ \frac{Δ x_{A}}{Δ x_{B}} = 2 \end{matrix}$

AP-EAMCET-2006

Structure of Atom

238800 If the uncertainty in velocity of a moving object is $1.0 \times 10^{- 6} {ms}^{- 1}$ and the uncertainty in its position is $58 m$ , The mass of this object is approximately equal to that of $(h = 6.626 \times 10^{- 34} Js)$

1 helium

2 deuterium

3 lithium

4 electron

Explanation:

Given-
Uncertainty in velocity $(Δ v) = 1.0 \times 10^{- 6} {ms}^{- 1}$
Uncertainty in position $(Δ x) = 58 m$
$Mass (m) = ?$
Now, from the Heisenberg principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
where $- Δ x =$ uncertainty in position $Δ p =$ uncertainty in momentum $∴ Δ x . Δ v = \frac{h}{4 π m}$ or $m = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{- 6}}$ 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 $0.1 kg$ has an uncertainty of $0.15 {ms}^{- 1}$ in its velocity then the uncertainty in the position of the cricket balls is

1

3.5 \times 10^{- 33} m

2

35 \times 10^{- 33} m

3

1.7 \times 10^{- 34} m

4

1.7 \times 10^{- 33} m

Explanation:

According to Heisenberg's Uncertainty principle-
$\Rightarrow Δ p . Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant $(6.67 \times 10^{- 34} J - s)$
$\begin{aligned} ∵ Δ p = m Δ v \\ \Rightarrow m Δ v Δ x \geq \frac{h}{4 π} \end{aligned}$
Given, $m = 0.1 kg, Δ x = 0.15 m / \sec$
$\begin{aligned} Δ x = \frac{h}{4 π m Δ v} \\ = \frac{6.67 \times 10^{- 34}}{4 \times 3.14 \times 0.1 \times 0.15} \\ = \frac{6.67}{0.1884} \times 10^{- 34} \\ \Rightarrow 35.40 \times 10^{- 34} \\ \Rightarrow 3.5 \times 10^{- 33} m \end{aligned}$

AP EAPCET-6 Sep. 2021

Structure of Atom

238803 Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^{4} cm . s^{- 1}$ . accurate up to $0.0016 %$ . (Given, $m_{e} = 9.1 \times 10^{- 28}$ g. $h = 6.626 \times 10^{- 27}$ erg.s)

1

1.211 mm

2

2.99 \times 10^{- 10} mm

3

0.121 mm

4

12.11 mm

Explanation:

Given -
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1}, m_{e} = 9.1 \times 10^{- 28} gm, \\ h = 6.626 \times 10^{- 27} g {cm}^{2} s^{- 2} \end{aligned}$
Hence,
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1} \times \frac{0.0016}{100} \\ Δ v = 0.478 cm s^{- 1} \end{aligned}$
From equation of Heisenberg uncertainity principle -
$Δ x . Δ v \geq \frac{h}{4 π m}$
Where, $h =$ Planck's constant
$Δ x =$ Uncertainity in position
$Δ v =$ Uncertainity in velocity
$m =$ Mass
$π = 3.14$
or $Δ x = \frac{6.626 \times 10^{- 27} (g {cm}^{2} s^{- 2}) \cdot s}{4 \times 3.14 \times 0.478 (cm s^{- 1}) \times 9.1 \times 10^{- 28} (gm)}$
$Δ x = 1.21 cm = 12.1 mm$

AP EAPCET 24.08.2021

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 ${ms}^{- 1})$ is about $(h =$ $6.63 \times 10^{- 34} Js)$

1

8 \times 10^{3}

2

8 \times 10^{4}

3

8 \times 10^{5}

4

8 \times 10^{6}

5

8 \times 10^{7}

Explanation:

$\begin{aligned} G i v e n t h a t - \\ M a s s o f p a r t i c l e $ (m) = 6.63 \times 10^{- 28} g \\ = 6.63 \times 10^{- 31} kg \\ P l a n c k^{'} s c o n s t a n t (h) = 6.63 \times 10^{- 34} Js \\ Δ x = 1 A^{\circ} = 10^{- 10} m \end{aligned}$
According to Heisenberg's uncertainty principle
$\begin{aligned} Δ x . Δ p = \frac{h}{4 π} \\ Δ x . m Δ v = \frac{h}{4 π} \\ Δ v = \frac{h}{4 π \times m \times Δ x} \\ Δ v = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 6.63 \times 10^{- 31} \times 10^{- 10}} \\ Δ v = \frac{10^{- 34} \times 10^{41}}{12.56} \\ Δ v = \frac{10^{7}}{12.56} \\ Δ v = \frac{100 \times 10^{5}}{12.56} \\ Δ v = 8 \times 10^{5} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238799 The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 {ms}^{- 1}$ respectively. The mass of $B$ is five times to that of $A$ . What is the ratio of uncertainties $(\frac{Δ x_{A}}{Δ x_{B}})$ their positions in :

1 2

2 0.25

3 4

4 1

Explanation:

Given, $Δ v_{A} = 0.05 {ms}^{- 1}$ and $m_{A} = m$
$Δ v_{B} = 0.02 {ms}^{- 1} m_{B} = 5 m$
According to Heisenberg uncertainty principle-
$m Δ v \cdot Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant
$\begin{aligned} Δ x = Uncertainity in position \\ Δ v = Uncertainity in velocity \\ m = Mass \end{aligned}$
For particle A :
$m \times 0.05 \times Δ x_{A} = \frac{h}{4 π}$
For particle B :
$5 m \times 0.02 \times Δ x_{B} = \frac{h}{4 π}$
Dividing equation (i) and (ii) we get :
$\begin{matrix} \frac{m \times 0.05 \times Δ x_{A}}{5 m \times 0.02 \times Δ x_{B}} = 1 \\ \frac{Δ x_{A}}{Δ x_{B}} = \frac{0.02 \times 5}{0.05} \\ \frac{Δ x_{A}}{Δ x_{B}} = 2 \end{matrix}$

AP-EAMCET-2006

Structure of Atom

238800 If the uncertainty in velocity of a moving object is $1.0 \times 10^{- 6} {ms}^{- 1}$ and the uncertainty in its position is $58 m$ , The mass of this object is approximately equal to that of $(h = 6.626 \times 10^{- 34} Js)$

1 helium

2 deuterium

3 lithium

4 electron

Explanation:

Given-
Uncertainty in velocity $(Δ v) = 1.0 \times 10^{- 6} {ms}^{- 1}$
Uncertainty in position $(Δ x) = 58 m$
$Mass (m) = ?$
Now, from the Heisenberg principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
where $- Δ x =$ uncertainty in position $Δ p =$ uncertainty in momentum $∴ Δ x . Δ v = \frac{h}{4 π m}$ or $m = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{- 6}}$ 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 $0.1 kg$ has an uncertainty of $0.15 {ms}^{- 1}$ in its velocity then the uncertainty in the position of the cricket balls is

1

3.5 \times 10^{- 33} m

2

35 \times 10^{- 33} m

3

1.7 \times 10^{- 34} m

4

1.7 \times 10^{- 33} m

Explanation:

According to Heisenberg's Uncertainty principle-
$\Rightarrow Δ p . Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant $(6.67 \times 10^{- 34} J - s)$
$\begin{aligned} ∵ Δ p = m Δ v \\ \Rightarrow m Δ v Δ x \geq \frac{h}{4 π} \end{aligned}$
Given, $m = 0.1 kg, Δ x = 0.15 m / \sec$
$\begin{aligned} Δ x = \frac{h}{4 π m Δ v} \\ = \frac{6.67 \times 10^{- 34}}{4 \times 3.14 \times 0.1 \times 0.15} \\ = \frac{6.67}{0.1884} \times 10^{- 34} \\ \Rightarrow 35.40 \times 10^{- 34} \\ \Rightarrow 3.5 \times 10^{- 33} m \end{aligned}$

AP EAPCET-6 Sep. 2021

Structure of Atom

238803 Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^{4} cm . s^{- 1}$ . accurate up to $0.0016 %$ . (Given, $m_{e} = 9.1 \times 10^{- 28}$ g. $h = 6.626 \times 10^{- 27}$ erg.s)

1

1.211 mm

2

2.99 \times 10^{- 10} mm

3

0.121 mm

4

12.11 mm

Explanation:

Given -
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1}, m_{e} = 9.1 \times 10^{- 28} gm, \\ h = 6.626 \times 10^{- 27} g {cm}^{2} s^{- 2} \end{aligned}$
Hence,
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1} \times \frac{0.0016}{100} \\ Δ v = 0.478 cm s^{- 1} \end{aligned}$
From equation of Heisenberg uncertainity principle -
$Δ x . Δ v \geq \frac{h}{4 π m}$
Where, $h =$ Planck's constant
$Δ x =$ Uncertainity in position
$Δ v =$ Uncertainity in velocity
$m =$ Mass
$π = 3.14$
or $Δ x = \frac{6.626 \times 10^{- 27} (g {cm}^{2} s^{- 2}) \cdot s}{4 \times 3.14 \times 0.478 (cm s^{- 1}) \times 9.1 \times 10^{- 28} (gm)}$
$Δ x = 1.21 cm = 12.1 mm$

AP EAPCET 24.08.2021

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 ${ms}^{- 1})$ is about $(h =$ $6.63 \times 10^{- 34} Js)$

1

8 \times 10^{3}

2

8 \times 10^{4}

3

8 \times 10^{5}

4

8 \times 10^{6}

5

8 \times 10^{7}

Explanation:

$\begin{aligned} G i v e n t h a t - \\ M a s s o f p a r t i c l e $ (m) = 6.63 \times 10^{- 28} g \\ = 6.63 \times 10^{- 31} kg \\ P l a n c k^{'} s c o n s t a n t (h) = 6.63 \times 10^{- 34} Js \\ Δ x = 1 A^{\circ} = 10^{- 10} m \end{aligned}$
According to Heisenberg's uncertainty principle
$\begin{aligned} Δ x . Δ p = \frac{h}{4 π} \\ Δ x . m Δ v = \frac{h}{4 π} \\ Δ v = \frac{h}{4 π \times m \times Δ x} \\ Δ v = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 6.63 \times 10^{- 31} \times 10^{- 10}} \\ Δ v = \frac{10^{- 34} \times 10^{41}}{12.56} \\ Δ v = \frac{10^{7}}{12.56} \\ Δ v = \frac{100 \times 10^{5}}{12.56} \\ Δ v = 8 \times 10^{5} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238799 The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 {ms}^{- 1}$ respectively. The mass of $B$ is five times to that of $A$ . What is the ratio of uncertainties $(\frac{Δ x_{A}}{Δ x_{B}})$ their positions in :

1 2

2 0.25

3 4

4 1

Explanation:

Given, $Δ v_{A} = 0.05 {ms}^{- 1}$ and $m_{A} = m$
$Δ v_{B} = 0.02 {ms}^{- 1} m_{B} = 5 m$
According to Heisenberg uncertainty principle-
$m Δ v \cdot Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant
$\begin{aligned} Δ x = Uncertainity in position \\ Δ v = Uncertainity in velocity \\ m = Mass \end{aligned}$
For particle A :
$m \times 0.05 \times Δ x_{A} = \frac{h}{4 π}$
For particle B :
$5 m \times 0.02 \times Δ x_{B} = \frac{h}{4 π}$
Dividing equation (i) and (ii) we get :
$\begin{matrix} \frac{m \times 0.05 \times Δ x_{A}}{5 m \times 0.02 \times Δ x_{B}} = 1 \\ \frac{Δ x_{A}}{Δ x_{B}} = \frac{0.02 \times 5}{0.05} \\ \frac{Δ x_{A}}{Δ x_{B}} = 2 \end{matrix}$

AP-EAMCET-2006

Structure of Atom

238800 If the uncertainty in velocity of a moving object is $1.0 \times 10^{- 6} {ms}^{- 1}$ and the uncertainty in its position is $58 m$ , The mass of this object is approximately equal to that of $(h = 6.626 \times 10^{- 34} Js)$

1 helium

2 deuterium

3 lithium

4 electron

Explanation:

Given-
Uncertainty in velocity $(Δ v) = 1.0 \times 10^{- 6} {ms}^{- 1}$
Uncertainty in position $(Δ x) = 58 m$
$Mass (m) = ?$
Now, from the Heisenberg principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
where $- Δ x =$ uncertainty in position $Δ p =$ uncertainty in momentum $∴ Δ x . Δ v = \frac{h}{4 π m}$ or $m = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{- 6}}$ 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 $0.1 kg$ has an uncertainty of $0.15 {ms}^{- 1}$ in its velocity then the uncertainty in the position of the cricket balls is

1

3.5 \times 10^{- 33} m

2

35 \times 10^{- 33} m

3

1.7 \times 10^{- 34} m

4

1.7 \times 10^{- 33} m

Explanation:

According to Heisenberg's Uncertainty principle-
$\Rightarrow Δ p . Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant $(6.67 \times 10^{- 34} J - s)$
$\begin{aligned} ∵ Δ p = m Δ v \\ \Rightarrow m Δ v Δ x \geq \frac{h}{4 π} \end{aligned}$
Given, $m = 0.1 kg, Δ x = 0.15 m / \sec$
$\begin{aligned} Δ x = \frac{h}{4 π m Δ v} \\ = \frac{6.67 \times 10^{- 34}}{4 \times 3.14 \times 0.1 \times 0.15} \\ = \frac{6.67}{0.1884} \times 10^{- 34} \\ \Rightarrow 35.40 \times 10^{- 34} \\ \Rightarrow 3.5 \times 10^{- 33} m \end{aligned}$

AP EAPCET-6 Sep. 2021

Structure of Atom

238803 Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^{4} cm . s^{- 1}$ . accurate up to $0.0016 %$ . (Given, $m_{e} = 9.1 \times 10^{- 28}$ g. $h = 6.626 \times 10^{- 27}$ erg.s)

1

1.211 mm

2

2.99 \times 10^{- 10} mm

3

0.121 mm

4

12.11 mm

Explanation:

Given -
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1}, m_{e} = 9.1 \times 10^{- 28} gm, \\ h = 6.626 \times 10^{- 27} g {cm}^{2} s^{- 2} \end{aligned}$
Hence,
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1} \times \frac{0.0016}{100} \\ Δ v = 0.478 cm s^{- 1} \end{aligned}$
From equation of Heisenberg uncertainity principle -
$Δ x . Δ v \geq \frac{h}{4 π m}$
Where, $h =$ Planck's constant
$Δ x =$ Uncertainity in position
$Δ v =$ Uncertainity in velocity
$m =$ Mass
$π = 3.14$
or $Δ x = \frac{6.626 \times 10^{- 27} (g {cm}^{2} s^{- 2}) \cdot s}{4 \times 3.14 \times 0.478 (cm s^{- 1}) \times 9.1 \times 10^{- 28} (gm)}$
$Δ x = 1.21 cm = 12.1 mm$

AP EAPCET 24.08.2021

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 ${ms}^{- 1})$ is about $(h =$ $6.63 \times 10^{- 34} Js)$

1

8 \times 10^{3}

2

8 \times 10^{4}

3

8 \times 10^{5}

4

8 \times 10^{6}

5

8 \times 10^{7}

Explanation:

$\begin{aligned} G i v e n t h a t - \\ M a s s o f p a r t i c l e $ (m) = 6.63 \times 10^{- 28} g \\ = 6.63 \times 10^{- 31} kg \\ P l a n c k^{'} s c o n s t a n t (h) = 6.63 \times 10^{- 34} Js \\ Δ x = 1 A^{\circ} = 10^{- 10} m \end{aligned}$
According to Heisenberg's uncertainty principle
$\begin{aligned} Δ x . Δ p = \frac{h}{4 π} \\ Δ x . m Δ v = \frac{h}{4 π} \\ Δ v = \frac{h}{4 π \times m \times Δ x} \\ Δ v = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 6.63 \times 10^{- 31} \times 10^{- 10}} \\ Δ v = \frac{10^{- 34} \times 10^{41}}{12.56} \\ Δ v = \frac{10^{7}}{12.56} \\ Δ v = \frac{100 \times 10^{5}}{12.56} \\ Δ v = 8 \times 10^{5} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238799 The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 {ms}^{- 1}$ respectively. The mass of $B$ is five times to that of $A$ . What is the ratio of uncertainties $(\frac{Δ x_{A}}{Δ x_{B}})$ their positions in :

1 2

2 0.25

3 4

4 1

Explanation:

Given, $Δ v_{A} = 0.05 {ms}^{- 1}$ and $m_{A} = m$
$Δ v_{B} = 0.02 {ms}^{- 1} m_{B} = 5 m$
According to Heisenberg uncertainty principle-
$m Δ v \cdot Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant
$\begin{aligned} Δ x = Uncertainity in position \\ Δ v = Uncertainity in velocity \\ m = Mass \end{aligned}$
For particle A :
$m \times 0.05 \times Δ x_{A} = \frac{h}{4 π}$
For particle B :
$5 m \times 0.02 \times Δ x_{B} = \frac{h}{4 π}$
Dividing equation (i) and (ii) we get :
$\begin{matrix} \frac{m \times 0.05 \times Δ x_{A}}{5 m \times 0.02 \times Δ x_{B}} = 1 \\ \frac{Δ x_{A}}{Δ x_{B}} = \frac{0.02 \times 5}{0.05} \\ \frac{Δ x_{A}}{Δ x_{B}} = 2 \end{matrix}$

AP-EAMCET-2006

Structure of Atom

238800 If the uncertainty in velocity of a moving object is $1.0 \times 10^{- 6} {ms}^{- 1}$ and the uncertainty in its position is $58 m$ , The mass of this object is approximately equal to that of $(h = 6.626 \times 10^{- 34} Js)$

1 helium

2 deuterium

3 lithium

4 electron

Explanation:

Given-
Uncertainty in velocity $(Δ v) = 1.0 \times 10^{- 6} {ms}^{- 1}$
Uncertainty in position $(Δ x) = 58 m$
$Mass (m) = ?$
Now, from the Heisenberg principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
where $- Δ x =$ uncertainty in position $Δ p =$ uncertainty in momentum $∴ Δ x . Δ v = \frac{h}{4 π m}$ or $m = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{- 6}}$ 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 $0.1 kg$ has an uncertainty of $0.15 {ms}^{- 1}$ in its velocity then the uncertainty in the position of the cricket balls is

1

3.5 \times 10^{- 33} m

2

35 \times 10^{- 33} m

3

1.7 \times 10^{- 34} m

4

1.7 \times 10^{- 33} m

Explanation:

According to Heisenberg's Uncertainty principle-
$\Rightarrow Δ p . Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant $(6.67 \times 10^{- 34} J - s)$
$\begin{aligned} ∵ Δ p = m Δ v \\ \Rightarrow m Δ v Δ x \geq \frac{h}{4 π} \end{aligned}$
Given, $m = 0.1 kg, Δ x = 0.15 m / \sec$
$\begin{aligned} Δ x = \frac{h}{4 π m Δ v} \\ = \frac{6.67 \times 10^{- 34}}{4 \times 3.14 \times 0.1 \times 0.15} \\ = \frac{6.67}{0.1884} \times 10^{- 34} \\ \Rightarrow 35.40 \times 10^{- 34} \\ \Rightarrow 3.5 \times 10^{- 33} m \end{aligned}$

AP EAPCET-6 Sep. 2021

Structure of Atom

238803 Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^{4} cm . s^{- 1}$ . accurate up to $0.0016 %$ . (Given, $m_{e} = 9.1 \times 10^{- 28}$ g. $h = 6.626 \times 10^{- 27}$ erg.s)

1

1.211 mm

2

2.99 \times 10^{- 10} mm

3

0.121 mm

4

12.11 mm

Explanation:

Given -
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1}, m_{e} = 9.1 \times 10^{- 28} gm, \\ h = 6.626 \times 10^{- 27} g {cm}^{2} s^{- 2} \end{aligned}$
Hence,
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1} \times \frac{0.0016}{100} \\ Δ v = 0.478 cm s^{- 1} \end{aligned}$
From equation of Heisenberg uncertainity principle -
$Δ x . Δ v \geq \frac{h}{4 π m}$
Where, $h =$ Planck's constant
$Δ x =$ Uncertainity in position
$Δ v =$ Uncertainity in velocity
$m =$ Mass
$π = 3.14$
or $Δ x = \frac{6.626 \times 10^{- 27} (g {cm}^{2} s^{- 2}) \cdot s}{4 \times 3.14 \times 0.478 (cm s^{- 1}) \times 9.1 \times 10^{- 28} (gm)}$
$Δ x = 1.21 cm = 12.1 mm$

AP EAPCET 24.08.2021

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 ${ms}^{- 1})$ is about $(h =$ $6.63 \times 10^{- 34} Js)$

1

8 \times 10^{3}

2

8 \times 10^{4}

3

8 \times 10^{5}

4

8 \times 10^{6}

5

8 \times 10^{7}

Explanation:

$\begin{aligned} G i v e n t h a t - \\ M a s s o f p a r t i c l e $ (m) = 6.63 \times 10^{- 28} g \\ = 6.63 \times 10^{- 31} kg \\ P l a n c k^{'} s c o n s t a n t (h) = 6.63 \times 10^{- 34} Js \\ Δ x = 1 A^{\circ} = 10^{- 10} m \end{aligned}$
According to Heisenberg's uncertainty principle
$\begin{aligned} Δ x . Δ p = \frac{h}{4 π} \\ Δ x . m Δ v = \frac{h}{4 π} \\ Δ v = \frac{h}{4 π \times m \times Δ x} \\ Δ v = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 6.63 \times 10^{- 31} \times 10^{- 10}} \\ Δ v = \frac{10^{- 34} \times 10^{41}}{12.56} \\ Δ v = \frac{10^{7}}{12.56} \\ Δ v = \frac{100 \times 10^{5}}{12.56} \\ Δ v = 8 \times 10^{5} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238799 The uncertainties in the velocities of two particles $A$ and $B$ are 0.05 and $0.02 {ms}^{- 1}$ respectively. The mass of $B$ is five times to that of $A$ . What is the ratio of uncertainties $(\frac{Δ x_{A}}{Δ x_{B}})$ their positions in :

1 2

2 0.25

3 4

4 1

Explanation:

Given, $Δ v_{A} = 0.05 {ms}^{- 1}$ and $m_{A} = m$
$Δ v_{B} = 0.02 {ms}^{- 1} m_{B} = 5 m$
According to Heisenberg uncertainty principle-
$m Δ v \cdot Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant
$\begin{aligned} Δ x = Uncertainity in position \\ Δ v = Uncertainity in velocity \\ m = Mass \end{aligned}$
For particle A :
$m \times 0.05 \times Δ x_{A} = \frac{h}{4 π}$
For particle B :
$5 m \times 0.02 \times Δ x_{B} = \frac{h}{4 π}$
Dividing equation (i) and (ii) we get :
$\begin{matrix} \frac{m \times 0.05 \times Δ x_{A}}{5 m \times 0.02 \times Δ x_{B}} = 1 \\ \frac{Δ x_{A}}{Δ x_{B}} = \frac{0.02 \times 5}{0.05} \\ \frac{Δ x_{A}}{Δ x_{B}} = 2 \end{matrix}$

AP-EAMCET-2006

Structure of Atom

238800 If the uncertainty in velocity of a moving object is $1.0 \times 10^{- 6} {ms}^{- 1}$ and the uncertainty in its position is $58 m$ , The mass of this object is approximately equal to that of $(h = 6.626 \times 10^{- 34} Js)$

1 helium

2 deuterium

3 lithium

4 electron

Explanation:

Given-
Uncertainty in velocity $(Δ v) = 1.0 \times 10^{- 6} {ms}^{- 1}$
Uncertainty in position $(Δ x) = 58 m$
$Mass (m) = ?$
Now, from the Heisenberg principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
where $- Δ x =$ uncertainty in position $Δ p =$ uncertainty in momentum $∴ Δ x . Δ v = \frac{h}{4 π m}$ or $m = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 58 \times 1.0 \times 10^{- 6}}$ 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 $0.1 kg$ has an uncertainty of $0.15 {ms}^{- 1}$ in its velocity then the uncertainty in the position of the cricket balls is

1

3.5 \times 10^{- 33} m

2

35 \times 10^{- 33} m

3

1.7 \times 10^{- 34} m

4

1.7 \times 10^{- 33} m

Explanation:

According to Heisenberg's Uncertainty principle-
$\Rightarrow Δ p . Δ x \geq \frac{h}{4 π}$
Where, $h =$ Planck's constant $(6.67 \times 10^{- 34} J - s)$
$\begin{aligned} ∵ Δ p = m Δ v \\ \Rightarrow m Δ v Δ x \geq \frac{h}{4 π} \end{aligned}$
Given, $m = 0.1 kg, Δ x = 0.15 m / \sec$
$\begin{aligned} Δ x = \frac{h}{4 π m Δ v} \\ = \frac{6.67 \times 10^{- 34}}{4 \times 3.14 \times 0.1 \times 0.15} \\ = \frac{6.67}{0.1884} \times 10^{- 34} \\ \Rightarrow 35.40 \times 10^{- 34} \\ \Rightarrow 3.5 \times 10^{- 33} m \end{aligned}$

AP EAPCET-6 Sep. 2021

Structure of Atom

238803 Find the uncertainlity in the position of an electron which is moving with a velocity of 2.99 $\times 10^{4} cm . s^{- 1}$ . accurate up to $0.0016 %$ . (Given, $m_{e} = 9.1 \times 10^{- 28}$ g. $h = 6.626 \times 10^{- 27}$ erg.s)

1

1.211 mm

2

2.99 \times 10^{- 10} mm

3

0.121 mm

4

12.11 mm

Explanation:

Given -
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1}, m_{e} = 9.1 \times 10^{- 28} gm, \\ h = 6.626 \times 10^{- 27} g {cm}^{2} s^{- 2} \end{aligned}$
Hence,
$\begin{aligned} Δ v = 2.99 \times 10^{4} cm s^{- 1} \times \frac{0.0016}{100} \\ Δ v = 0.478 cm s^{- 1} \end{aligned}$
From equation of Heisenberg uncertainity principle -
$Δ x . Δ v \geq \frac{h}{4 π m}$
Where, $h =$ Planck's constant
$Δ x =$ Uncertainity in position
$Δ v =$ Uncertainity in velocity
$m =$ Mass
$π = 3.14$
or $Δ x = \frac{6.626 \times 10^{- 27} (g {cm}^{2} s^{- 2}) \cdot s}{4 \times 3.14 \times 0.478 (cm s^{- 1}) \times 9.1 \times 10^{- 28} (gm)}$
$Δ x = 1.21 cm = 12.1 mm$

AP EAPCET 24.08.2021