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CHXI02:STRUCTURE OF ATOM

307409 How many photons of light having a wavelength of $4000 \overset{^{\circ}}{A}$ are necessary to provide 1 J of energy? $(h = 6 .63 \times 1 0^{- 34} J . s, c = 3 \times 1 0^{8} m / s)$

1

2 .01 \times 1 0^{18}

2

4 .97 \times 1 0^{18}

3

1 0^{18}

4

6 \times 1 0^{23}

Explanation:

Energy of a photon $= h v = \frac{h c}{λ}$
$= \frac{(6 .63 \times 1 0^{- 34} J . s) (3 \times 1 0^{8} m / s)}{(4000 \times 1 0^{- 10} m)}$
$= 4 .97 \times 1 0^{- 19} J .$
Number of photons required
$= \frac{1 J}{4 .97 \times 1 0^{- 19} J} = 2 .01 \times 1 0^{18}$

CHXI02:STRUCTURE OF ATOM

307410 A near $UV$ photon of $300 nm$ is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength $760 nm$ . Hence, wavelength of the second photon is

1

1060 nm

2

496 nm

3

300 nm

4

215 nm

Explanation:

Energy values are additives.
$E = E_{1} + E_{2}$
$\Rightarrow \frac{hc}{λ} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}} [∵ E = hv = \frac{hc}{λ}]$
or $\frac{1}{λ} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$ or $\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$
or $\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760} \Rightarrow λ_{2} = \frac{300 \times 760}{460}$
$\Rightarrow λ_{2} = 495.6 nm \approx 496 nm$

CHXI02:STRUCTURE OF ATOM

307411 The energies $E_{1}$ and $E_{2}$ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., $λ_{1}$ and $λ_{2}$ will be

1

λ = λ_{2}

2

λ_{1} = 2 λ_{2}

3

λ_{1} = \frac{1}{2} λ_{2}

4

λ_{1} = 4 λ_{2}

Explanation:

$E_{1} = \frac{h c}{λ_{1}}$ and $E_{2} = \frac{h c}{λ_{2}}$
$∴ \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{25}{50} = \frac{λ_{2}}{λ_{1}}$
$∴ λ_{1} = 2 λ_{2}$

CHXI02:STRUCTURE OF ATOM

307412 A bulb emits electromagnetic radiation of $660 nm$ wavelength. The total energy of radiation is $3 \times 10^{- 18} J$ . The number of emitted photons will be
$(h = 6.6 \times 10^{- 34} J \times s, c = 3 \times 10^{8} m / s)$

1 1

2 10

3 100

4 1000

Explanation:

Let, number of emitted photons $= n$
$∴ E = n h v = n \frac{h c}{λ}$
where, $v =$ frequency, $c =$ velocity of light, $λ =$ wavelength
$∴ n = \frac{E \times λ}{h c}$
$n = \frac{3 \times 1 0^{- 18} J \times 660 \times 1 0^{- 9} m}{626 \times 1 0^{- 34} J S \times 3 \times 1 0^{8} m s^{- 1}} = 10 p h o t o n s$

AIIMS - 2019

CHXI02:STRUCTURE OF ATOM

307409 How many photons of light having a wavelength of $4000 \overset{^{\circ}}{A}$ are necessary to provide 1 J of energy? $(h = 6 .63 \times 1 0^{- 34} J . s, c = 3 \times 1 0^{8} m / s)$

1

2 .01 \times 1 0^{18}

2

4 .97 \times 1 0^{18}

3

1 0^{18}

4

6 \times 1 0^{23}

Explanation:

Energy of a photon $= h v = \frac{h c}{λ}$
$= \frac{(6 .63 \times 1 0^{- 34} J . s) (3 \times 1 0^{8} m / s)}{(4000 \times 1 0^{- 10} m)}$
$= 4 .97 \times 1 0^{- 19} J .$
Number of photons required
$= \frac{1 J}{4 .97 \times 1 0^{- 19} J} = 2 .01 \times 1 0^{18}$

CHXI02:STRUCTURE OF ATOM

307410 A near $UV$ photon of $300 nm$ is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength $760 nm$ . Hence, wavelength of the second photon is

1

1060 nm

2

496 nm

3

300 nm

4

215 nm

Explanation:

Energy values are additives.
$E = E_{1} + E_{2}$
$\Rightarrow \frac{hc}{λ} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}} [∵ E = hv = \frac{hc}{λ}]$
or $\frac{1}{λ} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$ or $\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$
or $\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760} \Rightarrow λ_{2} = \frac{300 \times 760}{460}$
$\Rightarrow λ_{2} = 495.6 nm \approx 496 nm$

CHXI02:STRUCTURE OF ATOM

307411 The energies $E_{1}$ and $E_{2}$ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., $λ_{1}$ and $λ_{2}$ will be

1

λ = λ_{2}

2

λ_{1} = 2 λ_{2}

3

λ_{1} = \frac{1}{2} λ_{2}

4

λ_{1} = 4 λ_{2}

Explanation:

$E_{1} = \frac{h c}{λ_{1}}$ and $E_{2} = \frac{h c}{λ_{2}}$
$∴ \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{25}{50} = \frac{λ_{2}}{λ_{1}}$
$∴ λ_{1} = 2 λ_{2}$

CHXI02:STRUCTURE OF ATOM

307412 A bulb emits electromagnetic radiation of $660 nm$ wavelength. The total energy of radiation is $3 \times 10^{- 18} J$ . The number of emitted photons will be
$(h = 6.6 \times 10^{- 34} J \times s, c = 3 \times 10^{8} m / s)$

1 1

2 10

3 100

4 1000

Explanation:

Let, number of emitted photons $= n$
$∴ E = n h v = n \frac{h c}{λ}$
where, $v =$ frequency, $c =$ velocity of light, $λ =$ wavelength
$∴ n = \frac{E \times λ}{h c}$
$n = \frac{3 \times 1 0^{- 18} J \times 660 \times 1 0^{- 9} m}{626 \times 1 0^{- 34} J S \times 3 \times 1 0^{8} m s^{- 1}} = 10 p h o t o n s$

AIIMS - 2019

CHXI02:STRUCTURE OF ATOM

307409 How many photons of light having a wavelength of $4000 \overset{^{\circ}}{A}$ are necessary to provide 1 J of energy? $(h = 6 .63 \times 1 0^{- 34} J . s, c = 3 \times 1 0^{8} m / s)$

1

2 .01 \times 1 0^{18}

2

4 .97 \times 1 0^{18}

3

1 0^{18}

4

6 \times 1 0^{23}

Explanation:

Energy of a photon $= h v = \frac{h c}{λ}$
$= \frac{(6 .63 \times 1 0^{- 34} J . s) (3 \times 1 0^{8} m / s)}{(4000 \times 1 0^{- 10} m)}$
$= 4 .97 \times 1 0^{- 19} J .$
Number of photons required
$= \frac{1 J}{4 .97 \times 1 0^{- 19} J} = 2 .01 \times 1 0^{18}$

CHXI02:STRUCTURE OF ATOM

307410 A near $UV$ photon of $300 nm$ is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength $760 nm$ . Hence, wavelength of the second photon is

1

1060 nm

2

496 nm

3

300 nm

4

215 nm

Explanation:

Energy values are additives.
$E = E_{1} + E_{2}$
$\Rightarrow \frac{hc}{λ} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}} [∵ E = hv = \frac{hc}{λ}]$
or $\frac{1}{λ} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$ or $\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$
or $\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760} \Rightarrow λ_{2} = \frac{300 \times 760}{460}$
$\Rightarrow λ_{2} = 495.6 nm \approx 496 nm$

CHXI02:STRUCTURE OF ATOM

307411 The energies $E_{1}$ and $E_{2}$ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., $λ_{1}$ and $λ_{2}$ will be

1

λ = λ_{2}

2

λ_{1} = 2 λ_{2}

3

λ_{1} = \frac{1}{2} λ_{2}

4

λ_{1} = 4 λ_{2}

Explanation:

$E_{1} = \frac{h c}{λ_{1}}$ and $E_{2} = \frac{h c}{λ_{2}}$
$∴ \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{25}{50} = \frac{λ_{2}}{λ_{1}}$
$∴ λ_{1} = 2 λ_{2}$

CHXI02:STRUCTURE OF ATOM

307412 A bulb emits electromagnetic radiation of $660 nm$ wavelength. The total energy of radiation is $3 \times 10^{- 18} J$ . The number of emitted photons will be
$(h = 6.6 \times 10^{- 34} J \times s, c = 3 \times 10^{8} m / s)$

1 1

2 10

3 100

4 1000

Explanation:

Let, number of emitted photons $= n$
$∴ E = n h v = n \frac{h c}{λ}$
where, $v =$ frequency, $c =$ velocity of light, $λ =$ wavelength
$∴ n = \frac{E \times λ}{h c}$
$n = \frac{3 \times 1 0^{- 18} J \times 660 \times 1 0^{- 9} m}{626 \times 1 0^{- 34} J S \times 3 \times 1 0^{8} m s^{- 1}} = 10 p h o t o n s$

AIIMS - 2019

CHXI02:STRUCTURE OF ATOM

307409 How many photons of light having a wavelength of $4000 \overset{^{\circ}}{A}$ are necessary to provide 1 J of energy? $(h = 6 .63 \times 1 0^{- 34} J . s, c = 3 \times 1 0^{8} m / s)$

1

2 .01 \times 1 0^{18}

2

4 .97 \times 1 0^{18}

3

1 0^{18}

4

6 \times 1 0^{23}

Explanation:

Energy of a photon $= h v = \frac{h c}{λ}$
$= \frac{(6 .63 \times 1 0^{- 34} J . s) (3 \times 1 0^{8} m / s)}{(4000 \times 1 0^{- 10} m)}$
$= 4 .97 \times 1 0^{- 19} J .$
Number of photons required
$= \frac{1 J}{4 .97 \times 1 0^{- 19} J} = 2 .01 \times 1 0^{18}$

CHXI02:STRUCTURE OF ATOM

307410 A near $UV$ photon of $300 nm$ is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength $760 nm$ . Hence, wavelength of the second photon is

1

1060 nm

2

496 nm

3

300 nm

4

215 nm

Explanation:

Energy values are additives.
$E = E_{1} + E_{2}$
$\Rightarrow \frac{hc}{λ} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}} [∵ E = hv = \frac{hc}{λ}]$
or $\frac{1}{λ} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$ or $\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$
or $\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760} \Rightarrow λ_{2} = \frac{300 \times 760}{460}$
$\Rightarrow λ_{2} = 495.6 nm \approx 496 nm$

CHXI02:STRUCTURE OF ATOM

307411 The energies $E_{1}$ and $E_{2}$ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., $λ_{1}$ and $λ_{2}$ will be

1

λ = λ_{2}

2

λ_{1} = 2 λ_{2}

3

λ_{1} = \frac{1}{2} λ_{2}

4

λ_{1} = 4 λ_{2}

Explanation:

$E_{1} = \frac{h c}{λ_{1}}$ and $E_{2} = \frac{h c}{λ_{2}}$
$∴ \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{25}{50} = \frac{λ_{2}}{λ_{1}}$
$∴ λ_{1} = 2 λ_{2}$

CHXI02:STRUCTURE OF ATOM

307412 A bulb emits electromagnetic radiation of $660 nm$ wavelength. The total energy of radiation is $3 \times 10^{- 18} J$ . The number of emitted photons will be
$(h = 6.6 \times 10^{- 34} J \times s, c = 3 \times 10^{8} m / s)$

1 1

2 10

3 100

4 1000

Explanation:

Let, number of emitted photons $= n$
$∴ E = n h v = n \frac{h c}{λ}$
where, $v =$ frequency, $c =$ velocity of light, $λ =$ wavelength
$∴ n = \frac{E \times λ}{h c}$
$n = \frac{3 \times 1 0^{- 18} J \times 660 \times 1 0^{- 9} m}{626 \times 1 0^{- 34} J S \times 3 \times 1 0^{8} m s^{- 1}} = 10 p h o t o n s$

AIIMS - 2019

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