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CHXI02:STRUCTURE OF ATOM

307426 The dimension of Planck’s constant are same as

1 Energy

2 Angular momentum

3 Work

4 Power

Explanation:

Plank’s constant, h and angular momentum, mvr have same units kg $m^{2} s^{- 1}$

CHXI02:STRUCTURE OF ATOM

307427 Ionisation energy of gaseous Na atoms is $495 .5 k J m o l^{- 1}$ . The lowest possible frequency of light that ionises a sodium atom is $(h = 6 .626 \times 1 0^{- 34} J s, N_{A} = 6 .022 \times 1 0^{23} m o l^{- 1})$

1

7 .50 \times 1 0^{4} s^{- 1}

2

4 .76 \times 1 0^{4} s^{- 1}

3

3 .15 \times 1 0^{15} s^{- 1}

4

1 .24 \times 1 0^{15} s^{- 1}

Explanation:

Energy $= N_{A} h v$
$495 .5 = 6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34} \times v$
$v = \frac{495 .5 \times 1 0^{3} J}{6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34}} = 12 .4 \times 1 0^{14}$
$= 1 .24 \times 1 0^{15} s^{- 1}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307428 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. What is the ionization energy of sodium per atom?

1

494 .5 \times 1 0^{- 6} J / a t o m

2

3169 .5 \times 1 0^{- 10} J / a t o m

3

5 .85 \times 1 0^{- 15} J / a t o m

4

8 .207 \times 1 0^{- 19} J / a t o m

Explanation:

$E = \frac{h c}{λ}$
$= \frac{6 .62 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{242 \times 1 0^{- 9}} = 8 .207 \times 1 0^{- 19} J$

CHXI02:STRUCTURE OF ATOM

307429 Excited hydrogen atom emits light in the ultraviolet region at $2 .47 \times 1 0^{15} H z .$ With this frequency, the energy of a single photon is: $(h = 6 .63 \times 1 0^{- 34} J s)$

1

8 .041 \times 1 0^{- 40} J

2

2 .680 \times 1 0^{- 19} J

3

1 .640 \times 1 0^{- 18} J

4

6 .111 \times 1 0^{- 17} J

Explanation:

$E = h v$
$= 6 .63 \times 1 0^{- 34} \times 2 .47 \times 1 0^{15} = 1 .640 \times 1 0^{- 18} J$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307426 The dimension of Planck’s constant are same as

1 Energy

2 Angular momentum

3 Work

4 Power

Explanation:

Plank’s constant, h and angular momentum, mvr have same units kg $m^{2} s^{- 1}$

CHXI02:STRUCTURE OF ATOM

307427 Ionisation energy of gaseous Na atoms is $495 .5 k J m o l^{- 1}$ . The lowest possible frequency of light that ionises a sodium atom is $(h = 6 .626 \times 1 0^{- 34} J s, N_{A} = 6 .022 \times 1 0^{23} m o l^{- 1})$

1

7 .50 \times 1 0^{4} s^{- 1}

2

4 .76 \times 1 0^{4} s^{- 1}

3

3 .15 \times 1 0^{15} s^{- 1}

4

1 .24 \times 1 0^{15} s^{- 1}

Explanation:

Energy $= N_{A} h v$
$495 .5 = 6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34} \times v$
$v = \frac{495 .5 \times 1 0^{3} J}{6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34}} = 12 .4 \times 1 0^{14}$
$= 1 .24 \times 1 0^{15} s^{- 1}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307428 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. What is the ionization energy of sodium per atom?

1

494 .5 \times 1 0^{- 6} J / a t o m

2

3169 .5 \times 1 0^{- 10} J / a t o m

3

5 .85 \times 1 0^{- 15} J / a t o m

4

8 .207 \times 1 0^{- 19} J / a t o m

Explanation:

$E = \frac{h c}{λ}$
$= \frac{6 .62 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{242 \times 1 0^{- 9}} = 8 .207 \times 1 0^{- 19} J$

CHXI02:STRUCTURE OF ATOM

307429 Excited hydrogen atom emits light in the ultraviolet region at $2 .47 \times 1 0^{15} H z .$ With this frequency, the energy of a single photon is: $(h = 6 .63 \times 1 0^{- 34} J s)$

1

8 .041 \times 1 0^{- 40} J

2

2 .680 \times 1 0^{- 19} J

3

1 .640 \times 1 0^{- 18} J

4

6 .111 \times 1 0^{- 17} J

Explanation:

$E = h v$
$= 6 .63 \times 1 0^{- 34} \times 2 .47 \times 1 0^{15} = 1 .640 \times 1 0^{- 18} J$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307426 The dimension of Planck’s constant are same as

1 Energy

2 Angular momentum

3 Work

4 Power

Explanation:

Plank’s constant, h and angular momentum, mvr have same units kg $m^{2} s^{- 1}$

CHXI02:STRUCTURE OF ATOM

307427 Ionisation energy of gaseous Na atoms is $495 .5 k J m o l^{- 1}$ . The lowest possible frequency of light that ionises a sodium atom is $(h = 6 .626 \times 1 0^{- 34} J s, N_{A} = 6 .022 \times 1 0^{23} m o l^{- 1})$

1

7 .50 \times 1 0^{4} s^{- 1}

2

4 .76 \times 1 0^{4} s^{- 1}

3

3 .15 \times 1 0^{15} s^{- 1}

4

1 .24 \times 1 0^{15} s^{- 1}

Explanation:

Energy $= N_{A} h v$
$495 .5 = 6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34} \times v$
$v = \frac{495 .5 \times 1 0^{3} J}{6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34}} = 12 .4 \times 1 0^{14}$
$= 1 .24 \times 1 0^{15} s^{- 1}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307428 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. What is the ionization energy of sodium per atom?

1

494 .5 \times 1 0^{- 6} J / a t o m

2

3169 .5 \times 1 0^{- 10} J / a t o m

3

5 .85 \times 1 0^{- 15} J / a t o m

4

8 .207 \times 1 0^{- 19} J / a t o m

Explanation:

$E = \frac{h c}{λ}$
$= \frac{6 .62 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{242 \times 1 0^{- 9}} = 8 .207 \times 1 0^{- 19} J$

CHXI02:STRUCTURE OF ATOM

307429 Excited hydrogen atom emits light in the ultraviolet region at $2 .47 \times 1 0^{15} H z .$ With this frequency, the energy of a single photon is: $(h = 6 .63 \times 1 0^{- 34} J s)$

1

8 .041 \times 1 0^{- 40} J

2

2 .680 \times 1 0^{- 19} J

3

1 .640 \times 1 0^{- 18} J

4

6 .111 \times 1 0^{- 17} J

Explanation:

$E = h v$
$= 6 .63 \times 1 0^{- 34} \times 2 .47 \times 1 0^{15} = 1 .640 \times 1 0^{- 18} J$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307426 The dimension of Planck’s constant are same as

1 Energy

2 Angular momentum

3 Work

4 Power

Explanation:

Plank’s constant, h and angular momentum, mvr have same units kg $m^{2} s^{- 1}$

CHXI02:STRUCTURE OF ATOM

307427 Ionisation energy of gaseous Na atoms is $495 .5 k J m o l^{- 1}$ . The lowest possible frequency of light that ionises a sodium atom is $(h = 6 .626 \times 1 0^{- 34} J s, N_{A} = 6 .022 \times 1 0^{23} m o l^{- 1})$

1

7 .50 \times 1 0^{4} s^{- 1}

2

4 .76 \times 1 0^{4} s^{- 1}

3

3 .15 \times 1 0^{15} s^{- 1}

4

1 .24 \times 1 0^{15} s^{- 1}

Explanation:

Energy $= N_{A} h v$
$495 .5 = 6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34} \times v$
$v = \frac{495 .5 \times 1 0^{3} J}{6 .023 \times 1 0^{23} \times 6 .6 \times 1 0^{- 34}} = 12 .4 \times 1 0^{14}$
$= 1 .24 \times 1 0^{15} s^{- 1}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307428 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. What is the ionization energy of sodium per atom?

1

494 .5 \times 1 0^{- 6} J / a t o m

2

3169 .5 \times 1 0^{- 10} J / a t o m

3

5 .85 \times 1 0^{- 15} J / a t o m

4

8 .207 \times 1 0^{- 19} J / a t o m

Explanation:

$E = \frac{h c}{λ}$
$= \frac{6 .62 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{242 \times 1 0^{- 9}} = 8 .207 \times 1 0^{- 19} J$

CHXI02:STRUCTURE OF ATOM

307429 Excited hydrogen atom emits light in the ultraviolet region at $2 .47 \times 1 0^{15} H z .$ With this frequency, the energy of a single photon is: $(h = 6 .63 \times 1 0^{- 34} J s)$

1

8 .041 \times 1 0^{- 40} J

2

2 .680 \times 1 0^{- 19} J

3

1 .640 \times 1 0^{- 18} J

4

6 .111 \times 1 0^{- 17} J

Explanation:

$E = h v$
$= 6 .63 \times 1 0^{- 34} \times 2 .47 \times 1 0^{15} = 1 .640 \times 1 0^{- 18} J$

JEE - 2014