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CHXI02:STRUCTURE OF ATOM

307224 The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is:

1

6 .626 \times 1 0^{- 34} m

2

6 .626 \times 1 0^{- 38} m

3

6 .626 \times 1 0^{- 31} m

4

6 .626 \times 1 0^{- 30} m

Explanation:

$λ = \frac{h}{m v}$
$h = 6 .6 \times 1 0^{- 34} J s, m = 1000 k g$
$v = 36 k m / h r = \frac{36 \times 1 0^{3}}{60 \times 60} m / s e c = 10 m / s e c$
$∴ λ = \frac{6 .6 \times 1 0^{- 34}}{1 0^{3} \times 10} = 6 .6 \times 1 0^{- 38} m$

JEE - 2013

CHXI02:STRUCTURE OF ATOM

307225 If the de-Broglie wavelength of a particle of mass m is 100 times its velocity, then its value in terms of its mass (m) and Planck’s constant (h) is

1

\frac{1}{10} \sqrt{\frac{m}{h}}

2

10 \sqrt{\frac{h}{m}}

3

\frac{1}{10} \sqrt{\frac{h}{m}}

4

10 \sqrt{\frac{m}{h}}

Explanation:

Let wavelength of particle be x
So, velocity $(v) = \frac{x}{100}$
$λ = \frac{h}{m v}; x = \frac{h \times 100}{m \times x}$
$x^{2} = 100 \frac{h}{m} o r x = 10 \sqrt{\frac{h}{m}}$

CHXI02:STRUCTURE OF ATOM

307226 An electron is moving with the velocity equal to 10% of the velocity of light. Its de-Broglie wavelength will be

1

2 .4 \times 1 0^{- 12} c m

2

2 .4 \times 1 0^{- 18} c m

3

2 .4 \times 1 0^{- 9} c m

4

2 .4 \times 1 0^{- 15} c m

Explanation:

$λ = \frac{h}{m v} = \frac{6 .63 \times 1 0^{- 27} g c m^{2} s^{- 1}}{9 .1 \times 1 0^{- 28} g \times 3 \times 1 0^{9} c m}$
$= 2 .4 \times 1 0^{- 9} c m$

CHXI02:STRUCTURE OF ATOM

307227 A body of mass x kg is moving with a velocity of $100 m s^{- 1}$ . Its de Broglie wavelength is $6 .62 \times 1 0^{- 35} m$ .Hence x is $(h = 6 .62 \times 1 0^{- 34} J s)$

1

0 .25 k g

2

0 .15 k g

3

0 .2 k g

4

0 .1 k g

Explanation:

According to de Broglie’s equation,
$λ = \frac{h}{m v}$
Given, $λ = 6 .62 \times 1 0^{- 35} m, v = 100 m s^{- 1}$
$h = 6 .62 \times 10 - 34 J s a n d m = x k g$
$∴ 6 .62 \times 1 0^{- 35} = \frac{6 .62 \times 1 0^{- 34}}{x \times 100}$
$\Rightarrow x = 0 .1 k g$

KCET - 2007

CHXI02:STRUCTURE OF ATOM

307224 The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is:

1

6 .626 \times 1 0^{- 34} m

2

6 .626 \times 1 0^{- 38} m

3

6 .626 \times 1 0^{- 31} m

4

6 .626 \times 1 0^{- 30} m

Explanation:

$λ = \frac{h}{m v}$
$h = 6 .6 \times 1 0^{- 34} J s, m = 1000 k g$
$v = 36 k m / h r = \frac{36 \times 1 0^{3}}{60 \times 60} m / s e c = 10 m / s e c$
$∴ λ = \frac{6 .6 \times 1 0^{- 34}}{1 0^{3} \times 10} = 6 .6 \times 1 0^{- 38} m$

JEE - 2013

CHXI02:STRUCTURE OF ATOM

307225 If the de-Broglie wavelength of a particle of mass m is 100 times its velocity, then its value in terms of its mass (m) and Planck’s constant (h) is

1

\frac{1}{10} \sqrt{\frac{m}{h}}

2

10 \sqrt{\frac{h}{m}}

3

\frac{1}{10} \sqrt{\frac{h}{m}}

4

10 \sqrt{\frac{m}{h}}

Explanation:

Let wavelength of particle be x
So, velocity $(v) = \frac{x}{100}$
$λ = \frac{h}{m v}; x = \frac{h \times 100}{m \times x}$
$x^{2} = 100 \frac{h}{m} o r x = 10 \sqrt{\frac{h}{m}}$

CHXI02:STRUCTURE OF ATOM

307226 An electron is moving with the velocity equal to 10% of the velocity of light. Its de-Broglie wavelength will be

1

2 .4 \times 1 0^{- 12} c m

2

2 .4 \times 1 0^{- 18} c m

3

2 .4 \times 1 0^{- 9} c m

4

2 .4 \times 1 0^{- 15} c m

Explanation:

$λ = \frac{h}{m v} = \frac{6 .63 \times 1 0^{- 27} g c m^{2} s^{- 1}}{9 .1 \times 1 0^{- 28} g \times 3 \times 1 0^{9} c m}$
$= 2 .4 \times 1 0^{- 9} c m$

CHXI02:STRUCTURE OF ATOM

307227 A body of mass x kg is moving with a velocity of $100 m s^{- 1}$ . Its de Broglie wavelength is $6 .62 \times 1 0^{- 35} m$ .Hence x is $(h = 6 .62 \times 1 0^{- 34} J s)$

1

0 .25 k g

2

0 .15 k g

3

0 .2 k g

4

0 .1 k g

Explanation:

According to de Broglie’s equation,
$λ = \frac{h}{m v}$
Given, $λ = 6 .62 \times 1 0^{- 35} m, v = 100 m s^{- 1}$
$h = 6 .62 \times 10 - 34 J s a n d m = x k g$
$∴ 6 .62 \times 1 0^{- 35} = \frac{6 .62 \times 1 0^{- 34}}{x \times 100}$
$\Rightarrow x = 0 .1 k g$

KCET - 2007

CHXI02:STRUCTURE OF ATOM

307224 The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is:

1

6 .626 \times 1 0^{- 34} m

2

6 .626 \times 1 0^{- 38} m

3

6 .626 \times 1 0^{- 31} m

4

6 .626 \times 1 0^{- 30} m

Explanation:

$λ = \frac{h}{m v}$
$h = 6 .6 \times 1 0^{- 34} J s, m = 1000 k g$
$v = 36 k m / h r = \frac{36 \times 1 0^{3}}{60 \times 60} m / s e c = 10 m / s e c$
$∴ λ = \frac{6 .6 \times 1 0^{- 34}}{1 0^{3} \times 10} = 6 .6 \times 1 0^{- 38} m$

JEE - 2013

CHXI02:STRUCTURE OF ATOM

307225 If the de-Broglie wavelength of a particle of mass m is 100 times its velocity, then its value in terms of its mass (m) and Planck’s constant (h) is

1

\frac{1}{10} \sqrt{\frac{m}{h}}

2

10 \sqrt{\frac{h}{m}}

3

\frac{1}{10} \sqrt{\frac{h}{m}}

4

10 \sqrt{\frac{m}{h}}

Explanation:

Let wavelength of particle be x
So, velocity $(v) = \frac{x}{100}$
$λ = \frac{h}{m v}; x = \frac{h \times 100}{m \times x}$
$x^{2} = 100 \frac{h}{m} o r x = 10 \sqrt{\frac{h}{m}}$

CHXI02:STRUCTURE OF ATOM

307226 An electron is moving with the velocity equal to 10% of the velocity of light. Its de-Broglie wavelength will be

1

2 .4 \times 1 0^{- 12} c m

2

2 .4 \times 1 0^{- 18} c m

3

2 .4 \times 1 0^{- 9} c m

4

2 .4 \times 1 0^{- 15} c m

Explanation:

$λ = \frac{h}{m v} = \frac{6 .63 \times 1 0^{- 27} g c m^{2} s^{- 1}}{9 .1 \times 1 0^{- 28} g \times 3 \times 1 0^{9} c m}$
$= 2 .4 \times 1 0^{- 9} c m$

CHXI02:STRUCTURE OF ATOM

307227 A body of mass x kg is moving with a velocity of $100 m s^{- 1}$ . Its de Broglie wavelength is $6 .62 \times 1 0^{- 35} m$ .Hence x is $(h = 6 .62 \times 1 0^{- 34} J s)$

1

0 .25 k g

2

0 .15 k g

3

0 .2 k g

4

0 .1 k g

Explanation:

According to de Broglie’s equation,
$λ = \frac{h}{m v}$
Given, $λ = 6 .62 \times 1 0^{- 35} m, v = 100 m s^{- 1}$
$h = 6 .62 \times 10 - 34 J s a n d m = x k g$
$∴ 6 .62 \times 1 0^{- 35} = \frac{6 .62 \times 1 0^{- 34}}{x \times 100}$
$\Rightarrow x = 0 .1 k g$

KCET - 2007

CHXI02:STRUCTURE OF ATOM

307224 The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is:

1

6 .626 \times 1 0^{- 34} m

2

6 .626 \times 1 0^{- 38} m

3

6 .626 \times 1 0^{- 31} m

4

6 .626 \times 1 0^{- 30} m

Explanation:

$λ = \frac{h}{m v}$
$h = 6 .6 \times 1 0^{- 34} J s, m = 1000 k g$
$v = 36 k m / h r = \frac{36 \times 1 0^{3}}{60 \times 60} m / s e c = 10 m / s e c$
$∴ λ = \frac{6 .6 \times 1 0^{- 34}}{1 0^{3} \times 10} = 6 .6 \times 1 0^{- 38} m$

JEE - 2013

CHXI02:STRUCTURE OF ATOM

307225 If the de-Broglie wavelength of a particle of mass m is 100 times its velocity, then its value in terms of its mass (m) and Planck’s constant (h) is

1

\frac{1}{10} \sqrt{\frac{m}{h}}

2

10 \sqrt{\frac{h}{m}}

3

\frac{1}{10} \sqrt{\frac{h}{m}}

4

10 \sqrt{\frac{m}{h}}

Explanation:

Let wavelength of particle be x
So, velocity $(v) = \frac{x}{100}$
$λ = \frac{h}{m v}; x = \frac{h \times 100}{m \times x}$
$x^{2} = 100 \frac{h}{m} o r x = 10 \sqrt{\frac{h}{m}}$

CHXI02:STRUCTURE OF ATOM

307226 An electron is moving with the velocity equal to 10% of the velocity of light. Its de-Broglie wavelength will be

1

2 .4 \times 1 0^{- 12} c m

2

2 .4 \times 1 0^{- 18} c m

3

2 .4 \times 1 0^{- 9} c m

4

2 .4 \times 1 0^{- 15} c m

Explanation:

$λ = \frac{h}{m v} = \frac{6 .63 \times 1 0^{- 27} g c m^{2} s^{- 1}}{9 .1 \times 1 0^{- 28} g \times 3 \times 1 0^{9} c m}$
$= 2 .4 \times 1 0^{- 9} c m$

CHXI02:STRUCTURE OF ATOM

307227 A body of mass x kg is moving with a velocity of $100 m s^{- 1}$ . Its de Broglie wavelength is $6 .62 \times 1 0^{- 35} m$ .Hence x is $(h = 6 .62 \times 1 0^{- 34} J s)$

1

0 .25 k g

2

0 .15 k g

3

0 .2 k g

4

0 .1 k g

Explanation:

According to de Broglie’s equation,
$λ = \frac{h}{m v}$
Given, $λ = 6 .62 \times 1 0^{- 35} m, v = 100 m s^{- 1}$
$h = 6 .62 \times 10 - 34 J s a n d m = x k g$
$∴ 6 .62 \times 1 0^{- 35} = \frac{6 .62 \times 1 0^{- 34}}{x \times 100}$
$\Rightarrow x = 0 .1 k g$

KCET - 2007

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