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CHXI02:STRUCTURE OF ATOM

307413 What would be the ratio of energy of a photon of frequency $x s^{- 1}$ to $y s^{- 1}$ ?

1

x y

2

x / y

3

y / x

4

x + y

Explanation:

$E \propto v \Rightarrow \frac{E_{1}}{E_{2}} = \frac{v_{1}}{v_{2}} = \frac{x}{y}$

CHXI02:STRUCTURE OF ATOM

307414 A certain dye absorbs light of $λ = 4000 \overset{^{\circ}}{A}$ and then fluorescences light of $λ = 5000$ $A^{\circ}$
Assuming that under given conditions $50 %$ of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of the number of quanta emitted out to the number of quanta absorbed.

1

\frac{5}{8}

2

\frac{8}{5}

3

\frac{3}{8}

4

\frac{8}{3}

Explanation:

Let $n_{1}$ quanta of $4000$ $A^{\circ}$ are absorbed and $n_{2}$ quanta of $5000$ $A^{\circ}$ are re-emitted, and energy of flouroscence light $= \frac{1}{2} \times$ energy absorbed
$\Rightarrow n_{2} \frac{hc}{λ_{2}} = \frac{1}{2} \times n_{1} \frac{hc}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{λ_{2}} = \frac{1}{2} \frac{n_{1}}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{5000} = \frac{1}{2} \times \frac{n_{1}}{4000}$
$\Rightarrow \frac{n_{2}}{n_{1}} = \frac{5}{8}$

CHXI02:STRUCTURE OF ATOM

307415 The number of quanta of radiations of frequency $4 .75 \times 1 0^{13} s e c^{- 1}$ required to melt 100 g of ice is (The energy required to melt 1 g of ice is 350J)

1

1 0^{21}

2

1113 \times 1 0^{21}

3

6 \times 1 0^{23}

4

2 \times 1 0^{23}

Explanation:

$E = n h v$
$= n \times 6 .62 \times 1 0^{- 34} J s e c \times 4 .75 \times 1 0^{13} s e c^{- 1}$
$= n \times 31 .445 \times 1 0^{- 21} J$
Energy required to melt 100 g ice $= 350 J \times 100$
$= 35000 J$
$n \times 31 .445 \times 1 0^{- 21} = 35000$
$n = \frac{35000}{31 .445 \times 1 0^{- 21}} = 1113 \times 1 0^{21}$

CHXI02:STRUCTURE OF ATOM

307416 Number of photons having wavelength 632.8 nm, emitted by 5 mW laser source in 1 second is

1

1 .6 \times 1 0^{19}

2

1 .6 \times 1 0^{16}

3

1 .6 \times 1 0^{25}

4

1 .6 \times 1 0^{13}

Explanation:

$E = \frac{n h c}{λ}$
$5 \times 1 0^{- 3} J / s = \frac{n \times 1240 \times 1 .6 \times 1 0^{- 19}}{632 .8}$
$n = 1 .6 \times 1 0^{16}$

CHXI02:STRUCTURE OF ATOM

307413 What would be the ratio of energy of a photon of frequency $x s^{- 1}$ to $y s^{- 1}$ ?

1

x y

2

x / y

3

y / x

4

x + y

Explanation:

$E \propto v \Rightarrow \frac{E_{1}}{E_{2}} = \frac{v_{1}}{v_{2}} = \frac{x}{y}$

CHXI02:STRUCTURE OF ATOM

307414 A certain dye absorbs light of $λ = 4000 \overset{^{\circ}}{A}$ and then fluorescences light of $λ = 5000$ $A^{\circ}$
Assuming that under given conditions $50 %$ of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of the number of quanta emitted out to the number of quanta absorbed.

1

\frac{5}{8}

2

\frac{8}{5}

3

\frac{3}{8}

4

\frac{8}{3}

Explanation:

Let $n_{1}$ quanta of $4000$ $A^{\circ}$ are absorbed and $n_{2}$ quanta of $5000$ $A^{\circ}$ are re-emitted, and energy of flouroscence light $= \frac{1}{2} \times$ energy absorbed
$\Rightarrow n_{2} \frac{hc}{λ_{2}} = \frac{1}{2} \times n_{1} \frac{hc}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{λ_{2}} = \frac{1}{2} \frac{n_{1}}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{5000} = \frac{1}{2} \times \frac{n_{1}}{4000}$
$\Rightarrow \frac{n_{2}}{n_{1}} = \frac{5}{8}$

CHXI02:STRUCTURE OF ATOM

307415 The number of quanta of radiations of frequency $4 .75 \times 1 0^{13} s e c^{- 1}$ required to melt 100 g of ice is (The energy required to melt 1 g of ice is 350J)

1

1 0^{21}

2

1113 \times 1 0^{21}

3

6 \times 1 0^{23}

4

2 \times 1 0^{23}

Explanation:

$E = n h v$
$= n \times 6 .62 \times 1 0^{- 34} J s e c \times 4 .75 \times 1 0^{13} s e c^{- 1}$
$= n \times 31 .445 \times 1 0^{- 21} J$
Energy required to melt 100 g ice $= 350 J \times 100$
$= 35000 J$
$n \times 31 .445 \times 1 0^{- 21} = 35000$
$n = \frac{35000}{31 .445 \times 1 0^{- 21}} = 1113 \times 1 0^{21}$

CHXI02:STRUCTURE OF ATOM

307416 Number of photons having wavelength 632.8 nm, emitted by 5 mW laser source in 1 second is

1

1 .6 \times 1 0^{19}

2

1 .6 \times 1 0^{16}

3

1 .6 \times 1 0^{25}

4

1 .6 \times 1 0^{13}

Explanation:

$E = \frac{n h c}{λ}$
$5 \times 1 0^{- 3} J / s = \frac{n \times 1240 \times 1 .6 \times 1 0^{- 19}}{632 .8}$
$n = 1 .6 \times 1 0^{16}$

CHXI02:STRUCTURE OF ATOM

307413 What would be the ratio of energy of a photon of frequency $x s^{- 1}$ to $y s^{- 1}$ ?

1

x y

2

x / y

3

y / x

4

x + y

Explanation:

$E \propto v \Rightarrow \frac{E_{1}}{E_{2}} = \frac{v_{1}}{v_{2}} = \frac{x}{y}$

CHXI02:STRUCTURE OF ATOM

307414 A certain dye absorbs light of $λ = 4000 \overset{^{\circ}}{A}$ and then fluorescences light of $λ = 5000$ $A^{\circ}$
Assuming that under given conditions $50 %$ of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of the number of quanta emitted out to the number of quanta absorbed.

1

\frac{5}{8}

2

\frac{8}{5}

3

\frac{3}{8}

4

\frac{8}{3}

Explanation:

Let $n_{1}$ quanta of $4000$ $A^{\circ}$ are absorbed and $n_{2}$ quanta of $5000$ $A^{\circ}$ are re-emitted, and energy of flouroscence light $= \frac{1}{2} \times$ energy absorbed
$\Rightarrow n_{2} \frac{hc}{λ_{2}} = \frac{1}{2} \times n_{1} \frac{hc}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{λ_{2}} = \frac{1}{2} \frac{n_{1}}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{5000} = \frac{1}{2} \times \frac{n_{1}}{4000}$
$\Rightarrow \frac{n_{2}}{n_{1}} = \frac{5}{8}$

CHXI02:STRUCTURE OF ATOM

307415 The number of quanta of radiations of frequency $4 .75 \times 1 0^{13} s e c^{- 1}$ required to melt 100 g of ice is (The energy required to melt 1 g of ice is 350J)

1

1 0^{21}

2

1113 \times 1 0^{21}

3

6 \times 1 0^{23}

4

2 \times 1 0^{23}

Explanation:

$E = n h v$
$= n \times 6 .62 \times 1 0^{- 34} J s e c \times 4 .75 \times 1 0^{13} s e c^{- 1}$
$= n \times 31 .445 \times 1 0^{- 21} J$
Energy required to melt 100 g ice $= 350 J \times 100$
$= 35000 J$
$n \times 31 .445 \times 1 0^{- 21} = 35000$
$n = \frac{35000}{31 .445 \times 1 0^{- 21}} = 1113 \times 1 0^{21}$

CHXI02:STRUCTURE OF ATOM

307416 Number of photons having wavelength 632.8 nm, emitted by 5 mW laser source in 1 second is

1

1 .6 \times 1 0^{19}

2

1 .6 \times 1 0^{16}

3

1 .6 \times 1 0^{25}

4

1 .6 \times 1 0^{13}

Explanation:

$E = \frac{n h c}{λ}$
$5 \times 1 0^{- 3} J / s = \frac{n \times 1240 \times 1 .6 \times 1 0^{- 19}}{632 .8}$
$n = 1 .6 \times 1 0^{16}$

CHXI02:STRUCTURE OF ATOM

307413 What would be the ratio of energy of a photon of frequency $x s^{- 1}$ to $y s^{- 1}$ ?

1

x y

2

x / y

3

y / x

4

x + y

Explanation:

$E \propto v \Rightarrow \frac{E_{1}}{E_{2}} = \frac{v_{1}}{v_{2}} = \frac{x}{y}$

CHXI02:STRUCTURE OF ATOM

307414 A certain dye absorbs light of $λ = 4000 \overset{^{\circ}}{A}$ and then fluorescences light of $λ = 5000$ $A^{\circ}$
Assuming that under given conditions $50 %$ of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of the number of quanta emitted out to the number of quanta absorbed.

1

\frac{5}{8}

2

\frac{8}{5}

3

\frac{3}{8}

4

\frac{8}{3}

Explanation:

Let $n_{1}$ quanta of $4000$ $A^{\circ}$ are absorbed and $n_{2}$ quanta of $5000$ $A^{\circ}$ are re-emitted, and energy of flouroscence light $= \frac{1}{2} \times$ energy absorbed
$\Rightarrow n_{2} \frac{hc}{λ_{2}} = \frac{1}{2} \times n_{1} \frac{hc}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{λ_{2}} = \frac{1}{2} \frac{n_{1}}{λ_{1}}$
$\Rightarrow \frac{n_{2}}{5000} = \frac{1}{2} \times \frac{n_{1}}{4000}$
$\Rightarrow \frac{n_{2}}{n_{1}} = \frac{5}{8}$

CHXI02:STRUCTURE OF ATOM

307415 The number of quanta of radiations of frequency $4 .75 \times 1 0^{13} s e c^{- 1}$ required to melt 100 g of ice is (The energy required to melt 1 g of ice is 350J)

1

1 0^{21}

2

1113 \times 1 0^{21}

3

6 \times 1 0^{23}

4

2 \times 1 0^{23}

Explanation:

$E = n h v$
$= n \times 6 .62 \times 1 0^{- 34} J s e c \times 4 .75 \times 1 0^{13} s e c^{- 1}$
$= n \times 31 .445 \times 1 0^{- 21} J$
Energy required to melt 100 g ice $= 350 J \times 100$
$= 35000 J$
$n \times 31 .445 \times 1 0^{- 21} = 35000$
$n = \frac{35000}{31 .445 \times 1 0^{- 21}} = 1113 \times 1 0^{21}$

CHXI02:STRUCTURE OF ATOM

307416 Number of photons having wavelength 632.8 nm, emitted by 5 mW laser source in 1 second is

1

1 .6 \times 1 0^{19}

2

1 .6 \times 1 0^{16}

3

1 .6 \times 1 0^{25}

4

1 .6 \times 1 0^{13}

Explanation:

$E = \frac{n h c}{λ}$
$5 \times 1 0^{- 3} J / s = \frac{n \times 1240 \times 1 .6 \times 1 0^{- 19}}{632 .8}$
$n = 1 .6 \times 1 0^{16}$

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