QBManager

CHXI02:STRUCTURE OF ATOM

307385 Photoelectric effect can be explained by assuming that light

1 Is a form of transverse waves

2 Is a form of longitudinal waves

3 Can be polarised

4 Consists of quanta

CHXI02:STRUCTURE OF ATOM

307386 Ejection of the photoelectron from metal in the photoelectric effect experiment can by stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

1

4 e V

2

5 .5 e V

3

4 .5 e V

4

5 e V

Explanation:

$λ = 250 n m = 2500 \overset{o}{A}$
$E = \frac{h c}{λ} = \frac{12400}{2500} = 4 .96 e V$
$K E = s t o p p i n g p o t e n t i a l = 0 .5 e V$
$E = W_{0} + K . E$
$4 .96 = W + 0 .5$
$W_{0} = 4 .46 \approx 4 .5 e V$

JEE - 2018

CHXI02:STRUCTURE OF ATOM

307387 The energy absorbed by each molecule $(A_{2})$ of a substance is $4 .4 \times 1 0^{- 19} J$ and bond energy(BE) per molecule is $4 .0 \times 1 0^{- 19} J$ . The kinetic energy of the molecule per atom will be

1

2 .0 \times 1 0^{- 20} J

2

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

3

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

4

4 .0 \times 1 0^{- 20} J

Explanation:

KE of molecule = energy absorbed by molecule – BE per molecule
$= (4 .4 \times 1 0^{- 19}) - (4 .0 \times 1 0^{- 19}) J = 0 .4 \times 1 0^{- 19} J$
KE per atom $= \frac{0 .4 \times 1 0^{- 19}}{2} J = 2 .0 \times 1 0^{- 20} J$

CHXI02:STRUCTURE OF ATOM

307389 The work function of a photoelectric material is 3.3 eV, is threshold frequency will be

1

8 \times 1 0^{14} H z

2

5 \times 1 0^{33} H z

3

8 \times 1 0^{10} H z

4

4 \times 1 0^{11} H z

Explanation:

Threshold frequency
$(v_{o}) = \frac{w o r k f u n c t i o n}{h}$
$= \frac{3 .3 \times 1 .6 \times 1 0^{- 19} J}{6 .6 \times 1 0^{- 34} J s} = 8 \times 1 0^{14} H z$

CHXI02:STRUCTURE OF ATOM

307390 The potential to be applied to bring the photoelectron to rest which was emitted from a metal surface of work function 2.1 eV on irradiating it with radiation of frequency $6 \times 1 0^{14} H z$ is $(1 e V = 1 .6 \times 1 0^{- 19} J)$

1

0 .425 e V

2

0 .211 e V

3

2 .1 e V

4

0 .385 e V

Explanation:

Energy of incident photon (in eV)
$= h υ = \frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} = 2.485 e V$
K.E. $= h υ -$ Work function $= 2.485 - 2.1 = 0.385 e V$
Potential to be applied in order to stop the electron $= 0.385 e V$

CHXI02:STRUCTURE OF ATOM

307385 Photoelectric effect can be explained by assuming that light

1 Is a form of transverse waves

2 Is a form of longitudinal waves

3 Can be polarised

4 Consists of quanta

CHXI02:STRUCTURE OF ATOM

307386 Ejection of the photoelectron from metal in the photoelectric effect experiment can by stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

1

4 e V

2

5 .5 e V

3

4 .5 e V

4

5 e V

Explanation:

$λ = 250 n m = 2500 \overset{o}{A}$
$E = \frac{h c}{λ} = \frac{12400}{2500} = 4 .96 e V$
$K E = s t o p p i n g p o t e n t i a l = 0 .5 e V$
$E = W_{0} + K . E$
$4 .96 = W + 0 .5$
$W_{0} = 4 .46 \approx 4 .5 e V$

JEE - 2018

CHXI02:STRUCTURE OF ATOM

307387 The energy absorbed by each molecule $(A_{2})$ of a substance is $4 .4 \times 1 0^{- 19} J$ and bond energy(BE) per molecule is $4 .0 \times 1 0^{- 19} J$ . The kinetic energy of the molecule per atom will be

1

2 .0 \times 1 0^{- 20} J

2

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

3

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

4

4 .0 \times 1 0^{- 20} J

Explanation:

KE of molecule = energy absorbed by molecule – BE per molecule
$= (4 .4 \times 1 0^{- 19}) - (4 .0 \times 1 0^{- 19}) J = 0 .4 \times 1 0^{- 19} J$
KE per atom $= \frac{0 .4 \times 1 0^{- 19}}{2} J = 2 .0 \times 1 0^{- 20} J$

CHXI02:STRUCTURE OF ATOM

307389 The work function of a photoelectric material is 3.3 eV, is threshold frequency will be

1

8 \times 1 0^{14} H z

2

5 \times 1 0^{33} H z

3

8 \times 1 0^{10} H z

4

4 \times 1 0^{11} H z

Explanation:

Threshold frequency
$(v_{o}) = \frac{w o r k f u n c t i o n}{h}$
$= \frac{3 .3 \times 1 .6 \times 1 0^{- 19} J}{6 .6 \times 1 0^{- 34} J s} = 8 \times 1 0^{14} H z$

CHXI02:STRUCTURE OF ATOM

307390 The potential to be applied to bring the photoelectron to rest which was emitted from a metal surface of work function 2.1 eV on irradiating it with radiation of frequency $6 \times 1 0^{14} H z$ is $(1 e V = 1 .6 \times 1 0^{- 19} J)$

1

0 .425 e V

2

0 .211 e V

3

2 .1 e V

4

0 .385 e V

Explanation:

Energy of incident photon (in eV)
$= h υ = \frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} = 2.485 e V$
K.E. $= h υ -$ Work function $= 2.485 - 2.1 = 0.385 e V$
Potential to be applied in order to stop the electron $= 0.385 e V$

CHXI02:STRUCTURE OF ATOM

307385 Photoelectric effect can be explained by assuming that light

1 Is a form of transverse waves

2 Is a form of longitudinal waves

3 Can be polarised

4 Consists of quanta

CHXI02:STRUCTURE OF ATOM

307386 Ejection of the photoelectron from metal in the photoelectric effect experiment can by stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

1

4 e V

2

5 .5 e V

3

4 .5 e V

4

5 e V

Explanation:

$λ = 250 n m = 2500 \overset{o}{A}$
$E = \frac{h c}{λ} = \frac{12400}{2500} = 4 .96 e V$
$K E = s t o p p i n g p o t e n t i a l = 0 .5 e V$
$E = W_{0} + K . E$
$4 .96 = W + 0 .5$
$W_{0} = 4 .46 \approx 4 .5 e V$

JEE - 2018

CHXI02:STRUCTURE OF ATOM

307387 The energy absorbed by each molecule $(A_{2})$ of a substance is $4 .4 \times 1 0^{- 19} J$ and bond energy(BE) per molecule is $4 .0 \times 1 0^{- 19} J$ . The kinetic energy of the molecule per atom will be

1

2 .0 \times 1 0^{- 20} J

2

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

3

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

4

4 .0 \times 1 0^{- 20} J

Explanation:

KE of molecule = energy absorbed by molecule – BE per molecule
$= (4 .4 \times 1 0^{- 19}) - (4 .0 \times 1 0^{- 19}) J = 0 .4 \times 1 0^{- 19} J$
KE per atom $= \frac{0 .4 \times 1 0^{- 19}}{2} J = 2 .0 \times 1 0^{- 20} J$

CHXI02:STRUCTURE OF ATOM

307389 The work function of a photoelectric material is 3.3 eV, is threshold frequency will be

1

8 \times 1 0^{14} H z

2

5 \times 1 0^{33} H z

3

8 \times 1 0^{10} H z

4

4 \times 1 0^{11} H z

Explanation:

Threshold frequency
$(v_{o}) = \frac{w o r k f u n c t i o n}{h}$
$= \frac{3 .3 \times 1 .6 \times 1 0^{- 19} J}{6 .6 \times 1 0^{- 34} J s} = 8 \times 1 0^{14} H z$

CHXI02:STRUCTURE OF ATOM

307390 The potential to be applied to bring the photoelectron to rest which was emitted from a metal surface of work function 2.1 eV on irradiating it with radiation of frequency $6 \times 1 0^{14} H z$ is $(1 e V = 1 .6 \times 1 0^{- 19} J)$

1

0 .425 e V

2

0 .211 e V

3

2 .1 e V

4

0 .385 e V

Explanation:

Energy of incident photon (in eV)
$= h υ = \frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} = 2.485 e V$
K.E. $= h υ -$ Work function $= 2.485 - 2.1 = 0.385 e V$
Potential to be applied in order to stop the electron $= 0.385 e V$

CHXI02:STRUCTURE OF ATOM

307385 Photoelectric effect can be explained by assuming that light

1 Is a form of transverse waves

2 Is a form of longitudinal waves

3 Can be polarised

4 Consists of quanta

CHXI02:STRUCTURE OF ATOM

307386 Ejection of the photoelectron from metal in the photoelectric effect experiment can by stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

1

4 e V

2

5 .5 e V

3

4 .5 e V

4

5 e V

Explanation:

$λ = 250 n m = 2500 \overset{o}{A}$
$E = \frac{h c}{λ} = \frac{12400}{2500} = 4 .96 e V$
$K E = s t o p p i n g p o t e n t i a l = 0 .5 e V$
$E = W_{0} + K . E$
$4 .96 = W + 0 .5$
$W_{0} = 4 .46 \approx 4 .5 e V$

JEE - 2018

CHXI02:STRUCTURE OF ATOM

307387 The energy absorbed by each molecule $(A_{2})$ of a substance is $4 .4 \times 1 0^{- 19} J$ and bond energy(BE) per molecule is $4 .0 \times 1 0^{- 19} J$ . The kinetic energy of the molecule per atom will be

1

2 .0 \times 1 0^{- 20} J

2

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

3

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

4

4 .0 \times 1 0^{- 20} J

Explanation:

KE of molecule = energy absorbed by molecule – BE per molecule
$= (4 .4 \times 1 0^{- 19}) - (4 .0 \times 1 0^{- 19}) J = 0 .4 \times 1 0^{- 19} J$
KE per atom $= \frac{0 .4 \times 1 0^{- 19}}{2} J = 2 .0 \times 1 0^{- 20} J$

CHXI02:STRUCTURE OF ATOM

307389 The work function of a photoelectric material is 3.3 eV, is threshold frequency will be

1

8 \times 1 0^{14} H z

2

5 \times 1 0^{33} H z

3

8 \times 1 0^{10} H z

4

4 \times 1 0^{11} H z

Explanation:

Threshold frequency
$(v_{o}) = \frac{w o r k f u n c t i o n}{h}$
$= \frac{3 .3 \times 1 .6 \times 1 0^{- 19} J}{6 .6 \times 1 0^{- 34} J s} = 8 \times 1 0^{14} H z$

CHXI02:STRUCTURE OF ATOM

307390 The potential to be applied to bring the photoelectron to rest which was emitted from a metal surface of work function 2.1 eV on irradiating it with radiation of frequency $6 \times 1 0^{14} H z$ is $(1 e V = 1 .6 \times 1 0^{- 19} J)$

1

0 .425 e V

2

0 .211 e V

3

2 .1 e V

4

0 .385 e V

Explanation:

Energy of incident photon (in eV)
$= h υ = \frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} = 2.485 e V$
K.E. $= h υ -$ Work function $= 2.485 - 2.1 = 0.385 e V$
Potential to be applied in order to stop the electron $= 0.385 e V$

CHXI02:STRUCTURE OF ATOM

307385 Photoelectric effect can be explained by assuming that light

1 Is a form of transverse waves

2 Is a form of longitudinal waves

3 Can be polarised

4 Consists of quanta

CHXI02:STRUCTURE OF ATOM

307386 Ejection of the photoelectron from metal in the photoelectric effect experiment can by stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

1

4 e V

2

5 .5 e V

3

4 .5 e V

4

5 e V

Explanation:

$λ = 250 n m = 2500 \overset{o}{A}$
$E = \frac{h c}{λ} = \frac{12400}{2500} = 4 .96 e V$
$K E = s t o p p i n g p o t e n t i a l = 0 .5 e V$
$E = W_{0} + K . E$
$4 .96 = W + 0 .5$
$W_{0} = 4 .46 \approx 4 .5 e V$

JEE - 2018

CHXI02:STRUCTURE OF ATOM

307387 The energy absorbed by each molecule $(A_{2})$ of a substance is $4 .4 \times 1 0^{- 19} J$ and bond energy(BE) per molecule is $4 .0 \times 1 0^{- 19} J$ . The kinetic energy of the molecule per atom will be

1

2 .0 \times 1 0^{- 20} J

2

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

3

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

4

4 .0 \times 1 0^{- 20} J

Explanation:

KE of molecule = energy absorbed by molecule – BE per molecule
$= (4 .4 \times 1 0^{- 19}) - (4 .0 \times 1 0^{- 19}) J = 0 .4 \times 1 0^{- 19} J$
KE per atom $= \frac{0 .4 \times 1 0^{- 19}}{2} J = 2 .0 \times 1 0^{- 20} J$

CHXI02:STRUCTURE OF ATOM

307389 The work function of a photoelectric material is 3.3 eV, is threshold frequency will be

1

8 \times 1 0^{14} H z

2

5 \times 1 0^{33} H z

3

8 \times 1 0^{10} H z

4

4 \times 1 0^{11} H z

Explanation:

Threshold frequency
$(v_{o}) = \frac{w o r k f u n c t i o n}{h}$
$= \frac{3 .3 \times 1 .6 \times 1 0^{- 19} J}{6 .6 \times 1 0^{- 34} J s} = 8 \times 1 0^{14} H z$

CHXI02:STRUCTURE OF ATOM

307390 The potential to be applied to bring the photoelectron to rest which was emitted from a metal surface of work function 2.1 eV on irradiating it with radiation of frequency $6 \times 1 0^{14} H z$ is $(1 e V = 1 .6 \times 1 0^{- 19} J)$

1

0 .425 e V

2

0 .211 e V

3

2 .1 e V

4

0 .385 e V

Explanation:

Energy of incident photon (in eV)
$= h υ = \frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} = 2.485 e V$
K.E. $= h υ -$ Work function $= 2.485 - 2.1 = 0.385 e V$
Potential to be applied in order to stop the electron $= 0.385 e V$