QBManager

Structure of Atom

238806 Both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately. This is known as

1 De Broglie principle

2 Hamiltonian law

3 Heisenberg uncertainty principle

4 Bohr theory of hydrogen atom

Explanation:

Heisenberg uncertainty principle- It is impossible to determine the exact position and the exact momentum of the particle simultaneously.
According to Heisenberg uncertainty principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
Thus, lesser the error in the momentum more will be the error in the position of the particle.

TS-EAMCET 09.08.2021

Structure of Atom

238816 The measurement of the electron position if associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s s^{- 1}$ . The uncertainty in electron 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, $Δ p = m Δ v$
Substituting the given values of $Δ x$ and $m$ , we get $1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v$
or $Δ v = \frac{1 \times 10^{- 18}}{9 \times 10^{- 28}}$
$= 1.1 \times 10^{9} {cms}^{- 1} = 1 \times 10^{9} {cms}^{- 1}$

BITSAT 2007

Structure of Atom

238817 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

1

2 meV

2

\sqrt{meV}

3

\sqrt{2 meV}

4

meV

Explanation:

Wavelength, $λ = \frac{h}{mv} = \frac{h}{\sqrt{2 m (KE)}}$
$∴ \frac{h}{λ} = \sqrt{2 meV}$

[JEE Main-2016]

Structure of Atom

238835 The position of both, an electron and a helium atom is known within $1.0 nm$ . Further the momentum of the electron is known within 5.0 $\times 10^{- 26} kg m s^{- 1}$ . The minimum uncertainty in the measurement of the momentum of the helium stone is

1

8.0 \times 10^{- 26} kg m s^{- 1}

2

80 kg m s^{- 1}

3

50 kg m s^{- 1}

4

5.0 \times 10^{- 26} kg m s^{- 1}

Explanation:

Uncertainty principle state that the product of uncertainty position and uncertainty in momentum is constant for a particle
$Δ x, Δ p = \frac{h}{4 π}$
Here, given $Δ x = 1.0 nm$ for both electron and helium atom, so $Δ p$ charge of momentum is also same for both the particle
Therefore uncertainty in momentum of the helium stone is also $5.0 \times 10^{- 26} kg {ms}^{- 1}$

NEET-1998

Structure of Atom

238823 Heisenberg uncertainty principle can be explained as
#[Qdiff: Easy, QCat: Theory Based, examname: BCECE-2005
, JIPMER-2008
, JCECE - 2007]#

1

Δ x \geq \frac{Δ P \times h}{4 π}

2

Δ x \times Δ P \geq \frac{h}{4 π}

3

Δ x \times Δ P \geq \frac{h}{π}

4

Δ P \geq \frac{π h}{Δ x}

Explanation:

Exp:According to Heisenberg's uncertainty principle.
$Δ x . Δ p \geq \frac{h}{4 π} or Δ x . (m . Δ v) \geq \frac{h}{4 π}$

Structure of Atom

238806 Both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately. This is known as

1 De Broglie principle

2 Hamiltonian law

3 Heisenberg uncertainty principle

4 Bohr theory of hydrogen atom

Explanation:

Heisenberg uncertainty principle- It is impossible to determine the exact position and the exact momentum of the particle simultaneously.
According to Heisenberg uncertainty principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
Thus, lesser the error in the momentum more will be the error in the position of the particle.

TS-EAMCET 09.08.2021

Structure of Atom

238816 The measurement of the electron position if associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s s^{- 1}$ . The uncertainty in electron 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, $Δ p = m Δ v$
Substituting the given values of $Δ x$ and $m$ , we get $1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v$
or $Δ v = \frac{1 \times 10^{- 18}}{9 \times 10^{- 28}}$
$= 1.1 \times 10^{9} {cms}^{- 1} = 1 \times 10^{9} {cms}^{- 1}$

BITSAT 2007

Structure of Atom

238817 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

1

2 meV

2

\sqrt{meV}

3

\sqrt{2 meV}

4

meV

Explanation:

Wavelength, $λ = \frac{h}{mv} = \frac{h}{\sqrt{2 m (KE)}}$
$∴ \frac{h}{λ} = \sqrt{2 meV}$

[JEE Main-2016]

Structure of Atom

238835 The position of both, an electron and a helium atom is known within $1.0 nm$ . Further the momentum of the electron is known within 5.0 $\times 10^{- 26} kg m s^{- 1}$ . The minimum uncertainty in the measurement of the momentum of the helium stone is

1

8.0 \times 10^{- 26} kg m s^{- 1}

2

80 kg m s^{- 1}

3

50 kg m s^{- 1}

4

5.0 \times 10^{- 26} kg m s^{- 1}

Explanation:

Uncertainty principle state that the product of uncertainty position and uncertainty in momentum is constant for a particle
$Δ x, Δ p = \frac{h}{4 π}$
Here, given $Δ x = 1.0 nm$ for both electron and helium atom, so $Δ p$ charge of momentum is also same for both the particle
Therefore uncertainty in momentum of the helium stone is also $5.0 \times 10^{- 26} kg {ms}^{- 1}$

NEET-1998

Structure of Atom

238823 Heisenberg uncertainty principle can be explained as
#[Qdiff: Easy, QCat: Theory Based, examname: BCECE-2005
, JIPMER-2008
, JCECE - 2007]#

1

Δ x \geq \frac{Δ P \times h}{4 π}

2

Δ x \times Δ P \geq \frac{h}{4 π}

3

Δ x \times Δ P \geq \frac{h}{π}

4

Δ P \geq \frac{π h}{Δ x}

Explanation:

Exp:According to Heisenberg's uncertainty principle.
$Δ x . Δ p \geq \frac{h}{4 π} or Δ x . (m . Δ v) \geq \frac{h}{4 π}$

Structure of Atom

238806 Both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately. This is known as

1 De Broglie principle

2 Hamiltonian law

3 Heisenberg uncertainty principle

4 Bohr theory of hydrogen atom

Explanation:

Heisenberg uncertainty principle- It is impossible to determine the exact position and the exact momentum of the particle simultaneously.
According to Heisenberg uncertainty principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
Thus, lesser the error in the momentum more will be the error in the position of the particle.

TS-EAMCET 09.08.2021

Structure of Atom

238816 The measurement of the electron position if associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s s^{- 1}$ . The uncertainty in electron 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, $Δ p = m Δ v$
Substituting the given values of $Δ x$ and $m$ , we get $1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v$
or $Δ v = \frac{1 \times 10^{- 18}}{9 \times 10^{- 28}}$
$= 1.1 \times 10^{9} {cms}^{- 1} = 1 \times 10^{9} {cms}^{- 1}$

BITSAT 2007

Structure of Atom

238817 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

1

2 meV

2

\sqrt{meV}

3

\sqrt{2 meV}

4

meV

Explanation:

Wavelength, $λ = \frac{h}{mv} = \frac{h}{\sqrt{2 m (KE)}}$
$∴ \frac{h}{λ} = \sqrt{2 meV}$

[JEE Main-2016]

Structure of Atom

238835 The position of both, an electron and a helium atom is known within $1.0 nm$ . Further the momentum of the electron is known within 5.0 $\times 10^{- 26} kg m s^{- 1}$ . The minimum uncertainty in the measurement of the momentum of the helium stone is

1

8.0 \times 10^{- 26} kg m s^{- 1}

2

80 kg m s^{- 1}

3

50 kg m s^{- 1}

4

5.0 \times 10^{- 26} kg m s^{- 1}

Explanation:

Uncertainty principle state that the product of uncertainty position and uncertainty in momentum is constant for a particle
$Δ x, Δ p = \frac{h}{4 π}$
Here, given $Δ x = 1.0 nm$ for both electron and helium atom, so $Δ p$ charge of momentum is also same for both the particle
Therefore uncertainty in momentum of the helium stone is also $5.0 \times 10^{- 26} kg {ms}^{- 1}$

NEET-1998

Structure of Atom

238823 Heisenberg uncertainty principle can be explained as
#[Qdiff: Easy, QCat: Theory Based, examname: BCECE-2005
, JIPMER-2008
, JCECE - 2007]#

1

Δ x \geq \frac{Δ P \times h}{4 π}

2

Δ x \times Δ P \geq \frac{h}{4 π}

3

Δ x \times Δ P \geq \frac{h}{π}

4

Δ P \geq \frac{π h}{Δ x}

Explanation:

Exp:According to Heisenberg's uncertainty principle.
$Δ x . Δ p \geq \frac{h}{4 π} or Δ x . (m . Δ v) \geq \frac{h}{4 π}$

Structure of Atom

238806 Both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately. This is known as

1 De Broglie principle

2 Hamiltonian law

3 Heisenberg uncertainty principle

4 Bohr theory of hydrogen atom

Explanation:

Heisenberg uncertainty principle- It is impossible to determine the exact position and the exact momentum of the particle simultaneously.
According to Heisenberg uncertainty principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
Thus, lesser the error in the momentum more will be the error in the position of the particle.

TS-EAMCET 09.08.2021

Structure of Atom

238816 The measurement of the electron position if associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s s^{- 1}$ . The uncertainty in electron 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, $Δ p = m Δ v$
Substituting the given values of $Δ x$ and $m$ , we get $1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v$
or $Δ v = \frac{1 \times 10^{- 18}}{9 \times 10^{- 28}}$
$= 1.1 \times 10^{9} {cms}^{- 1} = 1 \times 10^{9} {cms}^{- 1}$

BITSAT 2007

Structure of Atom

238817 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

1

2 meV

2

\sqrt{meV}

3

\sqrt{2 meV}

4

meV

Explanation:

Wavelength, $λ = \frac{h}{mv} = \frac{h}{\sqrt{2 m (KE)}}$
$∴ \frac{h}{λ} = \sqrt{2 meV}$

[JEE Main-2016]

Structure of Atom

238835 The position of both, an electron and a helium atom is known within $1.0 nm$ . Further the momentum of the electron is known within 5.0 $\times 10^{- 26} kg m s^{- 1}$ . The minimum uncertainty in the measurement of the momentum of the helium stone is

1

8.0 \times 10^{- 26} kg m s^{- 1}

2

80 kg m s^{- 1}

3

50 kg m s^{- 1}

4

5.0 \times 10^{- 26} kg m s^{- 1}

Explanation:

Uncertainty principle state that the product of uncertainty position and uncertainty in momentum is constant for a particle
$Δ x, Δ p = \frac{h}{4 π}$
Here, given $Δ x = 1.0 nm$ for both electron and helium atom, so $Δ p$ charge of momentum is also same for both the particle
Therefore uncertainty in momentum of the helium stone is also $5.0 \times 10^{- 26} kg {ms}^{- 1}$

NEET-1998

Structure of Atom

238823 Heisenberg uncertainty principle can be explained as
#[Qdiff: Easy, QCat: Theory Based, examname: BCECE-2005
, JIPMER-2008
, JCECE - 2007]#

1

Δ x \geq \frac{Δ P \times h}{4 π}

2

Δ x \times Δ P \geq \frac{h}{4 π}

3

Δ x \times Δ P \geq \frac{h}{π}

4

Δ P \geq \frac{π h}{Δ x}

Explanation:

Exp:According to Heisenberg's uncertainty principle.
$Δ x . Δ p \geq \frac{h}{4 π} or Δ x . (m . Δ v) \geq \frac{h}{4 π}$

Structure of Atom

238806 Both the position and exact velocity of an electron in an atom cannot be determined simultaneously and accurately. This is known as

1 De Broglie principle

2 Hamiltonian law

3 Heisenberg uncertainty principle

4 Bohr theory of hydrogen atom

Explanation:

Heisenberg uncertainty principle- It is impossible to determine the exact position and the exact momentum of the particle simultaneously.
According to Heisenberg uncertainty principle-
$Δ x . Δ p \geq \frac{h}{4 π}$
Thus, lesser the error in the momentum more will be the error in the position of the particle.

TS-EAMCET 09.08.2021

Structure of Atom

238816 The measurement of the electron position if associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g cm s s^{- 1}$ . The uncertainty in electron 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, $Δ p = m Δ v$
Substituting the given values of $Δ x$ and $m$ , we get $1 \times 10^{- 18} g cm s^{- 1} = 9 \times 10^{- 28} g \times Δ v$
or $Δ v = \frac{1 \times 10^{- 18}}{9 \times 10^{- 28}}$
$= 1.1 \times 10^{9} {cms}^{- 1} = 1 \times 10^{9} {cms}^{- 1}$

BITSAT 2007

Structure of Atom

238817 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

1

2 meV

2

\sqrt{meV}

3

\sqrt{2 meV}

4

meV

Explanation:

Wavelength, $λ = \frac{h}{mv} = \frac{h}{\sqrt{2 m (KE)}}$
$∴ \frac{h}{λ} = \sqrt{2 meV}$

[JEE Main-2016]

Structure of Atom

238835 The position of both, an electron and a helium atom is known within $1.0 nm$ . Further the momentum of the electron is known within 5.0 $\times 10^{- 26} kg m s^{- 1}$ . The minimum uncertainty in the measurement of the momentum of the helium stone is

1

8.0 \times 10^{- 26} kg m s^{- 1}

2

80 kg m s^{- 1}

3

50 kg m s^{- 1}

4

5.0 \times 10^{- 26} kg m s^{- 1}

Explanation:

Uncertainty principle state that the product of uncertainty position and uncertainty in momentum is constant for a particle
$Δ x, Δ p = \frac{h}{4 π}$
Here, given $Δ x = 1.0 nm$ for both electron and helium atom, so $Δ p$ charge of momentum is also same for both the particle
Therefore uncertainty in momentum of the helium stone is also $5.0 \times 10^{- 26} kg {ms}^{- 1}$

NEET-1998

Structure of Atom

238823 Heisenberg uncertainty principle can be explained as
#[Qdiff: Easy, QCat: Theory Based, examname: BCECE-2005
, JIPMER-2008
, JCECE - 2007]#

1

Δ x \geq \frac{Δ P \times h}{4 π}

2

Δ x \times Δ P \geq \frac{h}{4 π}

3

Δ x \times Δ P \geq \frac{h}{π}

4

Δ P \geq \frac{π h}{Δ x}

Explanation:

Exp:According to Heisenberg's uncertainty principle.
$Δ x . Δ p \geq \frac{h}{4 π} or Δ x . (m . Δ v) \geq \frac{h}{4 π}$