QBManager

Structure of Atom

238828 If uncertainty in position and velocity are equal, then uncertainty in momentum will be :

1

\frac{1}{2} \sqrt{\frac{mh}{π}}

2

\frac{1}{2} \sqrt{\frac{h}{π m}}

3

\frac{h}{4 π m}

4

\frac{mh}{4 π}

Explanation:

Given, $Δ x = Δ v$
According to Heisenberg Uncertainity Principle-
$Δ x \times Δ p \geq \frac{h}{4 π}$
Where, $Δ x =$ Change in position
$Δ p =$ Change in momentum
$\begin{matrix} Δ x \times m Δ v = \frac{h}{4 π} (∵ Δ p = m Δ v) \\ (Δ v^{2}) = \frac{h}{4 π m} [∵ Δ x = Δ v] \\ Δ p = m Δ v = m \sqrt{\frac{h}{4 π m}} \Rightarrow \sqrt{\frac{mh}{4 π}} \\ Δ p = \frac{1}{2} \sqrt{\frac{mh}{π}} \end{matrix}$

Manipal-2018

Structure of Atom

238829 Uncertainty in the position of an electron (mass $= 9.1 \times 10^{- 31} kg$ ) moving with a velocity $300 {ms}^{- 1}$ accurate upon $0.001 %$ will be $(h = 6.63 \times 10^{- 34} J - s)$

1

19.2 \times 10^{- 2} m

2

5.76 \times 10^{- 2} m

3

1.92 \times 10^{- 2} m

4

3.84 \times 10^{- 2} m

Explanation:

$Δ v = \frac{0.001}{100} \times 300 {ms}^{- 1} = 3 \times 10^{- 3} {ms}^{- 1}$
$\begin{aligned} Δ x \cdot m Δ v = \frac{h}{4 π} \\ Δ x = \frac{h}{4 π m Δ v} \\ \Rightarrow \frac{6.63 \times 10^{- 34} kg m^{2} s^{- 1}}{4 \times 3.14 \times 9.1 \times 10^{- 31} kg \times 3 \times 10^{- 3} {ms}^{- 1}} \\ = 0.019336 \\ Δ x ≃ 1.933 \times 10^{- 2} m \end{aligned}$

MHT CET-2011

Structure of Atom

238831 If uncertainty in position and momentum are equal, then uncertainty in velocity is

1

\frac{1}{m} \sqrt{\frac{h}{π}}

2

\sqrt{\frac{h}{π}}

3

\frac{1}{2 m} \sqrt{\frac{h}{π}}

4

\sqrt{\frac{h}{2 π}}

Explanation:

According to Heisenberg,
Uncertainty Principle $- Δ x . Δ p = \frac{h}{4 π}$
$(Δ p)^{2} = \frac{h}{4 π} [∴ Δ x = Δ p]$
$\begin{aligned} (m \cdot Δ v)^{2} = \frac{h}{4 π} \\ m^{2} \cdot Δ v^{2} = \frac{h}{4 π} \end{aligned}$
$Δ v = \sqrt{\frac{h}{4 π m^{2}}} = \frac{1}{2 m} \sqrt{\frac{h}{π}}$

NEET-2008

Structure of Atom

238832 The measurement of the electron position is associated with uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s^{- 1}$ . The uncertainty in election velocity is (mass of an electron is $9 \times 10^{- 28} g$ )

1

1 \times 10^{9} cm s^{- 1}

2

1 \times 10^{6} cm s^{- 1}

3

1 \times 10^{5} cm s^{- 1}

4

1 \times 10^{11} cm s^{- 1}

Explanation:

Momentum, $Δ p = m Δ v$
$\begin{aligned} 1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v \\ ∴ Δ v = 1 \times 10^{9} {cms}^{- 1} \end{aligned}$

NEET-2008

Structure of Atom

238828 If uncertainty in position and velocity are equal, then uncertainty in momentum will be :

1

\frac{1}{2} \sqrt{\frac{mh}{π}}

2

\frac{1}{2} \sqrt{\frac{h}{π m}}

3

\frac{h}{4 π m}

4

\frac{mh}{4 π}

Explanation:

Given, $Δ x = Δ v$
According to Heisenberg Uncertainity Principle-
$Δ x \times Δ p \geq \frac{h}{4 π}$
Where, $Δ x =$ Change in position
$Δ p =$ Change in momentum
$\begin{matrix} Δ x \times m Δ v = \frac{h}{4 π} (∵ Δ p = m Δ v) \\ (Δ v^{2}) = \frac{h}{4 π m} [∵ Δ x = Δ v] \\ Δ p = m Δ v = m \sqrt{\frac{h}{4 π m}} \Rightarrow \sqrt{\frac{mh}{4 π}} \\ Δ p = \frac{1}{2} \sqrt{\frac{mh}{π}} \end{matrix}$

Manipal-2018

Structure of Atom

238829 Uncertainty in the position of an electron (mass $= 9.1 \times 10^{- 31} kg$ ) moving with a velocity $300 {ms}^{- 1}$ accurate upon $0.001 %$ will be $(h = 6.63 \times 10^{- 34} J - s)$

1

19.2 \times 10^{- 2} m

2

5.76 \times 10^{- 2} m

3

1.92 \times 10^{- 2} m

4

3.84 \times 10^{- 2} m

Explanation:

$Δ v = \frac{0.001}{100} \times 300 {ms}^{- 1} = 3 \times 10^{- 3} {ms}^{- 1}$
$\begin{aligned} Δ x \cdot m Δ v = \frac{h}{4 π} \\ Δ x = \frac{h}{4 π m Δ v} \\ \Rightarrow \frac{6.63 \times 10^{- 34} kg m^{2} s^{- 1}}{4 \times 3.14 \times 9.1 \times 10^{- 31} kg \times 3 \times 10^{- 3} {ms}^{- 1}} \\ = 0.019336 \\ Δ x ≃ 1.933 \times 10^{- 2} m \end{aligned}$

MHT CET-2011

Structure of Atom

238831 If uncertainty in position and momentum are equal, then uncertainty in velocity is

1

\frac{1}{m} \sqrt{\frac{h}{π}}

2

\sqrt{\frac{h}{π}}

3

\frac{1}{2 m} \sqrt{\frac{h}{π}}

4

\sqrt{\frac{h}{2 π}}

Explanation:

According to Heisenberg,
Uncertainty Principle $- Δ x . Δ p = \frac{h}{4 π}$
$(Δ p)^{2} = \frac{h}{4 π} [∴ Δ x = Δ p]$
$\begin{aligned} (m \cdot Δ v)^{2} = \frac{h}{4 π} \\ m^{2} \cdot Δ v^{2} = \frac{h}{4 π} \end{aligned}$
$Δ v = \sqrt{\frac{h}{4 π m^{2}}} = \frac{1}{2 m} \sqrt{\frac{h}{π}}$

NEET-2008

Structure of Atom

238832 The measurement of the electron position is associated with uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s^{- 1}$ . The uncertainty in election velocity is (mass of an electron is $9 \times 10^{- 28} g$ )

1

1 \times 10^{9} cm s^{- 1}

2

1 \times 10^{6} cm s^{- 1}

3

1 \times 10^{5} cm s^{- 1}

4

1 \times 10^{11} cm s^{- 1}

Explanation:

Momentum, $Δ p = m Δ v$
$\begin{aligned} 1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v \\ ∴ Δ v = 1 \times 10^{9} {cms}^{- 1} \end{aligned}$

NEET-2008

Structure of Atom

238828 If uncertainty in position and velocity are equal, then uncertainty in momentum will be :

1

\frac{1}{2} \sqrt{\frac{mh}{π}}

2

\frac{1}{2} \sqrt{\frac{h}{π m}}

3

\frac{h}{4 π m}

4

\frac{mh}{4 π}

Explanation:

Given, $Δ x = Δ v$
According to Heisenberg Uncertainity Principle-
$Δ x \times Δ p \geq \frac{h}{4 π}$
Where, $Δ x =$ Change in position
$Δ p =$ Change in momentum
$\begin{matrix} Δ x \times m Δ v = \frac{h}{4 π} (∵ Δ p = m Δ v) \\ (Δ v^{2}) = \frac{h}{4 π m} [∵ Δ x = Δ v] \\ Δ p = m Δ v = m \sqrt{\frac{h}{4 π m}} \Rightarrow \sqrt{\frac{mh}{4 π}} \\ Δ p = \frac{1}{2} \sqrt{\frac{mh}{π}} \end{matrix}$

Manipal-2018

Structure of Atom

238829 Uncertainty in the position of an electron (mass $= 9.1 \times 10^{- 31} kg$ ) moving with a velocity $300 {ms}^{- 1}$ accurate upon $0.001 %$ will be $(h = 6.63 \times 10^{- 34} J - s)$

1

19.2 \times 10^{- 2} m

2

5.76 \times 10^{- 2} m

3

1.92 \times 10^{- 2} m

4

3.84 \times 10^{- 2} m

Explanation:

$Δ v = \frac{0.001}{100} \times 300 {ms}^{- 1} = 3 \times 10^{- 3} {ms}^{- 1}$
$\begin{aligned} Δ x \cdot m Δ v = \frac{h}{4 π} \\ Δ x = \frac{h}{4 π m Δ v} \\ \Rightarrow \frac{6.63 \times 10^{- 34} kg m^{2} s^{- 1}}{4 \times 3.14 \times 9.1 \times 10^{- 31} kg \times 3 \times 10^{- 3} {ms}^{- 1}} \\ = 0.019336 \\ Δ x ≃ 1.933 \times 10^{- 2} m \end{aligned}$

MHT CET-2011

Structure of Atom

238831 If uncertainty in position and momentum are equal, then uncertainty in velocity is

1

\frac{1}{m} \sqrt{\frac{h}{π}}

2

\sqrt{\frac{h}{π}}

3

\frac{1}{2 m} \sqrt{\frac{h}{π}}

4

\sqrt{\frac{h}{2 π}}

Explanation:

According to Heisenberg,
Uncertainty Principle $- Δ x . Δ p = \frac{h}{4 π}$
$(Δ p)^{2} = \frac{h}{4 π} [∴ Δ x = Δ p]$
$\begin{aligned} (m \cdot Δ v)^{2} = \frac{h}{4 π} \\ m^{2} \cdot Δ v^{2} = \frac{h}{4 π} \end{aligned}$
$Δ v = \sqrt{\frac{h}{4 π m^{2}}} = \frac{1}{2 m} \sqrt{\frac{h}{π}}$

NEET-2008

Structure of Atom

238832 The measurement of the electron position is associated with uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s^{- 1}$ . The uncertainty in election velocity is (mass of an electron is $9 \times 10^{- 28} g$ )

1

1 \times 10^{9} cm s^{- 1}

2

1 \times 10^{6} cm s^{- 1}

3

1 \times 10^{5} cm s^{- 1}

4

1 \times 10^{11} cm s^{- 1}

Explanation:

Momentum, $Δ p = m Δ v$
$\begin{aligned} 1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v \\ ∴ Δ v = 1 \times 10^{9} {cms}^{- 1} \end{aligned}$

NEET-2008

Structure of Atom

238828 If uncertainty in position and velocity are equal, then uncertainty in momentum will be :

1

\frac{1}{2} \sqrt{\frac{mh}{π}}

2

\frac{1}{2} \sqrt{\frac{h}{π m}}

3

\frac{h}{4 π m}

4

\frac{mh}{4 π}

Explanation:

Given, $Δ x = Δ v$
According to Heisenberg Uncertainity Principle-
$Δ x \times Δ p \geq \frac{h}{4 π}$
Where, $Δ x =$ Change in position
$Δ p =$ Change in momentum
$\begin{matrix} Δ x \times m Δ v = \frac{h}{4 π} (∵ Δ p = m Δ v) \\ (Δ v^{2}) = \frac{h}{4 π m} [∵ Δ x = Δ v] \\ Δ p = m Δ v = m \sqrt{\frac{h}{4 π m}} \Rightarrow \sqrt{\frac{mh}{4 π}} \\ Δ p = \frac{1}{2} \sqrt{\frac{mh}{π}} \end{matrix}$

Manipal-2018

Structure of Atom

238829 Uncertainty in the position of an electron (mass $= 9.1 \times 10^{- 31} kg$ ) moving with a velocity $300 {ms}^{- 1}$ accurate upon $0.001 %$ will be $(h = 6.63 \times 10^{- 34} J - s)$

1

19.2 \times 10^{- 2} m

2

5.76 \times 10^{- 2} m

3

1.92 \times 10^{- 2} m

4

3.84 \times 10^{- 2} m

Explanation:

$Δ v = \frac{0.001}{100} \times 300 {ms}^{- 1} = 3 \times 10^{- 3} {ms}^{- 1}$
$\begin{aligned} Δ x \cdot m Δ v = \frac{h}{4 π} \\ Δ x = \frac{h}{4 π m Δ v} \\ \Rightarrow \frac{6.63 \times 10^{- 34} kg m^{2} s^{- 1}}{4 \times 3.14 \times 9.1 \times 10^{- 31} kg \times 3 \times 10^{- 3} {ms}^{- 1}} \\ = 0.019336 \\ Δ x ≃ 1.933 \times 10^{- 2} m \end{aligned}$

MHT CET-2011

Structure of Atom

238831 If uncertainty in position and momentum are equal, then uncertainty in velocity is

1

\frac{1}{m} \sqrt{\frac{h}{π}}

2

\sqrt{\frac{h}{π}}

3

\frac{1}{2 m} \sqrt{\frac{h}{π}}

4

\sqrt{\frac{h}{2 π}}

Explanation:

According to Heisenberg,
Uncertainty Principle $- Δ x . Δ p = \frac{h}{4 π}$
$(Δ p)^{2} = \frac{h}{4 π} [∴ Δ x = Δ p]$
$\begin{aligned} (m \cdot Δ v)^{2} = \frac{h}{4 π} \\ m^{2} \cdot Δ v^{2} = \frac{h}{4 π} \end{aligned}$
$Δ v = \sqrt{\frac{h}{4 π m^{2}}} = \frac{1}{2 m} \sqrt{\frac{h}{π}}$

NEET-2008

Structure of Atom

238832 The measurement of the electron position is associated with uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s^{- 1}$ . The uncertainty in election velocity is (mass of an electron is $9 \times 10^{- 28} g$ )

1

1 \times 10^{9} cm s^{- 1}

2

1 \times 10^{6} cm s^{- 1}

3

1 \times 10^{5} cm s^{- 1}

4

1 \times 10^{11} cm s^{- 1}

Explanation:

Momentum, $Δ p = m Δ v$
$\begin{aligned} 1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v \\ ∴ Δ v = 1 \times 10^{9} {cms}^{- 1} \end{aligned}$

NEET-2008