QBManager

Structure of Atom

238737 The wavelength of a 150 g rubber ball moving with a velocity of $50 {ms}^{- 1}$ is :

1

3.43 \times 10^{- 33} cm

2

5.86 \times 10^{- 33} cm

3

7.77 \times 10^{- 33} cm

4

8.83 \times 10^{- 33} cm

Explanation:

: Given that, $v = 50 m / \sec, m = 150 gm =$ $150 \times 10^{- 3} kg$
According to de-Broglie -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.626 \times 10^{- 34}}{(150 \times 10^{- 3}) \times 50} \\ \Rightarrow 8.83 \times 10^{- 35} m \\ \Rightarrow 8.83 \times 10^{- 33} cm \end{aligned}$

AIIMS-1998

Structure of Atom

238728 Assertion: Threshold frequency is the maximum frequency required for the ejection of electron from the metal surface.
Reason: Threshold frequency is characteristic of a metal.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-26 May

Structure of Atom

238738 The de-Broglie wavelength of a particle with mass $1 g$ and velocity $100 m / s$ is

1

6.63 \times 10^{- 35} m

2

6.63 \times 10^{- 34} m

3

6.63 \times 10^{- 33} m

4

6.65 \times 10^{- 35} m

Explanation:

: Given that, $m = 1 g, v = 100 m / s$ $h = 6.63 \times 10^{- 34} J - s$
According to de-Broglie wave equation -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.63 \times 10^{- 34} J - s}{1 \times 10^{- 3} Kg \times 100 m / s} \\ = 6.63 \times 10^{- 33} m \end{aligned}$

NEET-1999

Structure of Atom

238742 In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, $a_{0} = 52.9 pm$ ]

1

211.6 pm

2

211.6 π pm

3

52.9 π pm

4

105.8 pm

Explanation:

: According to Bohr,
$\begin{aligned} mvr = \frac{nh}{2 π} \\ 2 π r = \frac{nh}{mv} = n λ \end{aligned}$
Where, $r =$ Radius
$λ =$ Wavelength
$n =$ Number of orbit
Also, $r = \frac{a_{0} n^{2}}{z}$
Where, $a_{o} =$ Bohr radius $= 52.9 pm$
$Z =$ Atomic number
On substituting the value of ' $r$ ' from equation (ii) to equation (i) we get
$\begin{aligned} n λ = \frac{2 π n^{2} a_{o}}{z} \\ λ = \frac{2 π {na}_{o}}{z} \\ λ = 2 π \times 2 \times 52.9 [∵ n = 2, z = 1] \\ λ = 211.6 π pm \end{aligned}$

NEET-Odisha 2019

Structure of Atom

238737 The wavelength of a 150 g rubber ball moving with a velocity of $50 {ms}^{- 1}$ is :

1

3.43 \times 10^{- 33} cm

2

5.86 \times 10^{- 33} cm

3

7.77 \times 10^{- 33} cm

4

8.83 \times 10^{- 33} cm

Explanation:

: Given that, $v = 50 m / \sec, m = 150 gm =$ $150 \times 10^{- 3} kg$
According to de-Broglie -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.626 \times 10^{- 34}}{(150 \times 10^{- 3}) \times 50} \\ \Rightarrow 8.83 \times 10^{- 35} m \\ \Rightarrow 8.83 \times 10^{- 33} cm \end{aligned}$

AIIMS-1998

Structure of Atom

238728 Assertion: Threshold frequency is the maximum frequency required for the ejection of electron from the metal surface.
Reason: Threshold frequency is characteristic of a metal.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-26 May

Structure of Atom

238738 The de-Broglie wavelength of a particle with mass $1 g$ and velocity $100 m / s$ is

1

6.63 \times 10^{- 35} m

2

6.63 \times 10^{- 34} m

3

6.63 \times 10^{- 33} m

4

6.65 \times 10^{- 35} m

Explanation:

: Given that, $m = 1 g, v = 100 m / s$ $h = 6.63 \times 10^{- 34} J - s$
According to de-Broglie wave equation -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.63 \times 10^{- 34} J - s}{1 \times 10^{- 3} Kg \times 100 m / s} \\ = 6.63 \times 10^{- 33} m \end{aligned}$

NEET-1999

Structure of Atom

238742 In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, $a_{0} = 52.9 pm$ ]

1

211.6 pm

2

211.6 π pm

3

52.9 π pm

4

105.8 pm

Explanation:

: According to Bohr,
$\begin{aligned} mvr = \frac{nh}{2 π} \\ 2 π r = \frac{nh}{mv} = n λ \end{aligned}$
Where, $r =$ Radius
$λ =$ Wavelength
$n =$ Number of orbit
Also, $r = \frac{a_{0} n^{2}}{z}$
Where, $a_{o} =$ Bohr radius $= 52.9 pm$
$Z =$ Atomic number
On substituting the value of ' $r$ ' from equation (ii) to equation (i) we get
$\begin{aligned} n λ = \frac{2 π n^{2} a_{o}}{z} \\ λ = \frac{2 π {na}_{o}}{z} \\ λ = 2 π \times 2 \times 52.9 [∵ n = 2, z = 1] \\ λ = 211.6 π pm \end{aligned}$

NEET-Odisha 2019

Structure of Atom

238737 The wavelength of a 150 g rubber ball moving with a velocity of $50 {ms}^{- 1}$ is :

1

3.43 \times 10^{- 33} cm

2

5.86 \times 10^{- 33} cm

3

7.77 \times 10^{- 33} cm

4

8.83 \times 10^{- 33} cm

Explanation:

: Given that, $v = 50 m / \sec, m = 150 gm =$ $150 \times 10^{- 3} kg$
According to de-Broglie -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.626 \times 10^{- 34}}{(150 \times 10^{- 3}) \times 50} \\ \Rightarrow 8.83 \times 10^{- 35} m \\ \Rightarrow 8.83 \times 10^{- 33} cm \end{aligned}$

AIIMS-1998

Structure of Atom

238728 Assertion: Threshold frequency is the maximum frequency required for the ejection of electron from the metal surface.
Reason: Threshold frequency is characteristic of a metal.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-26 May

Structure of Atom

238738 The de-Broglie wavelength of a particle with mass $1 g$ and velocity $100 m / s$ is

1

6.63 \times 10^{- 35} m

2

6.63 \times 10^{- 34} m

3

6.63 \times 10^{- 33} m

4

6.65 \times 10^{- 35} m

Explanation:

: Given that, $m = 1 g, v = 100 m / s$ $h = 6.63 \times 10^{- 34} J - s$
According to de-Broglie wave equation -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.63 \times 10^{- 34} J - s}{1 \times 10^{- 3} Kg \times 100 m / s} \\ = 6.63 \times 10^{- 33} m \end{aligned}$

NEET-1999

Structure of Atom

238742 In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, $a_{0} = 52.9 pm$ ]

1

211.6 pm

2

211.6 π pm

3

52.9 π pm

4

105.8 pm

Explanation:

: According to Bohr,
$\begin{aligned} mvr = \frac{nh}{2 π} \\ 2 π r = \frac{nh}{mv} = n λ \end{aligned}$
Where, $r =$ Radius
$λ =$ Wavelength
$n =$ Number of orbit
Also, $r = \frac{a_{0} n^{2}}{z}$
Where, $a_{o} =$ Bohr radius $= 52.9 pm$
$Z =$ Atomic number
On substituting the value of ' $r$ ' from equation (ii) to equation (i) we get
$\begin{aligned} n λ = \frac{2 π n^{2} a_{o}}{z} \\ λ = \frac{2 π {na}_{o}}{z} \\ λ = 2 π \times 2 \times 52.9 [∵ n = 2, z = 1] \\ λ = 211.6 π pm \end{aligned}$

NEET-Odisha 2019

Structure of Atom

238737 The wavelength of a 150 g rubber ball moving with a velocity of $50 {ms}^{- 1}$ is :

1

3.43 \times 10^{- 33} cm

2

5.86 \times 10^{- 33} cm

3

7.77 \times 10^{- 33} cm

4

8.83 \times 10^{- 33} cm

Explanation:

: Given that, $v = 50 m / \sec, m = 150 gm =$ $150 \times 10^{- 3} kg$
According to de-Broglie -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.626 \times 10^{- 34}}{(150 \times 10^{- 3}) \times 50} \\ \Rightarrow 8.83 \times 10^{- 35} m \\ \Rightarrow 8.83 \times 10^{- 33} cm \end{aligned}$

AIIMS-1998

Structure of Atom

238728 Assertion: Threshold frequency is the maximum frequency required for the ejection of electron from the metal surface.
Reason: Threshold frequency is characteristic of a metal.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-26 May

Structure of Atom

238738 The de-Broglie wavelength of a particle with mass $1 g$ and velocity $100 m / s$ is

1

6.63 \times 10^{- 35} m

2

6.63 \times 10^{- 34} m

3

6.63 \times 10^{- 33} m

4

6.65 \times 10^{- 35} m

Explanation:

: Given that, $m = 1 g, v = 100 m / s$ $h = 6.63 \times 10^{- 34} J - s$
According to de-Broglie wave equation -
$\begin{aligned} λ = \frac{h}{mv} = \frac{6.63 \times 10^{- 34} J - s}{1 \times 10^{- 3} Kg \times 100 m / s} \\ = 6.63 \times 10^{- 33} m \end{aligned}$

NEET-1999

Structure of Atom

238742 In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, $a_{0} = 52.9 pm$ ]

1

211.6 pm

2

211.6 π pm

3

52.9 π pm

4

105.8 pm

Explanation:

: According to Bohr,
$\begin{aligned} mvr = \frac{nh}{2 π} \\ 2 π r = \frac{nh}{mv} = n λ \end{aligned}$
Where, $r =$ Radius
$λ =$ Wavelength
$n =$ Number of orbit
Also, $r = \frac{a_{0} n^{2}}{z}$
Where, $a_{o} =$ Bohr radius $= 52.9 pm$
$Z =$ Atomic number
On substituting the value of ' $r$ ' from equation (ii) to equation (i) we get
$\begin{aligned} n λ = \frac{2 π n^{2} a_{o}}{z} \\ λ = \frac{2 π {na}_{o}}{z} \\ λ = 2 π \times 2 \times 52.9 [∵ n = 2, z = 1] \\ λ = 211.6 π pm \end{aligned}$

NEET-Odisha 2019