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CHXI02:STRUCTURE OF ATOM

307403 Einstein’s photoelectric equation states that $E_{k} = h v - W$ . In this equation, $E_{k}$ refers to

1 Kinetic energy of all ejected electrons

2 Mean kinetic energy of emitted electrons

3 Minimum kinetic energy of emitted electrons

4 Maximum kinetic energy of emitted electrons

CHXI02:STRUCTURE OF ATOM

307404 Assertion :
The energy of quantum of radiation is given by $E = h v$ .
Reason :
Quantum, in the energy equation signifies the principal quantum number.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency $(v)$ and is expressed by the equation. $E = h v$ . So option (3) is correct.

CHXI02:STRUCTURE OF ATOM

307405 The energy of one mole of photons of radiation of wavelength 300nm is
(Given: $h = 6 .63 \times 1 0^{- 34} J s$ , $N_{A} = 6 .02 \times 1 0^{23} m o l^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ )

1

235 k J m o l^{- 1}

2

325 k J m o l^{- 1}

3

399 k J m o l^{- 1}

4

435 k J m o l^{- 1}

Explanation:

Energy of one mole of photons $= \frac{h c}{λ} \times N_{A}$
$= \frac{6 .63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}} \times 6 .02 \times 1 0^{23}$
$= 399 .13 \times 1 0^{3} J o u l e / m o l e$
$= 399 k J / J o u l e$

JEE - 2022

CHXI02:STRUCTURE OF ATOM

307406 The energy required to break 1 mol of Cl-Cl bonds in $C l_{2}$ is $242 k J / m o l .$ The longest wavelength of light capable of breaking a single Cl-Cl bond is

1

594 n m

2

640 n m

3

700 n m

4

494 n m

Explanation:

Energy required for 1 $C l_{2}$ molecule
$= \frac{242 \times 1 0^{3}}{N_{A}} J$
$E = \frac{h c}{λ}$ or $λ = \frac{h c}{E}$
$= \frac{6 .626 \times 1 0^{- 34} \times 3 \times 1 0^{8} \times 6 .02 \times 1 0^{23}}{242 \times 1 0^{3}}$
$= 494 \times 1 0^{- 9} m = 494 n m$

CHXI02:STRUCTURE OF ATOM

307407 Number of photons emitted by a $100 W ({Js}^{- 1})$ yellow lamp in 1.0s is ( $λ$ of yellow light is $560 nm$ )

1

1.6 \times 10^{18}

2

1.4 \times 10^{18}

3

2.8 \times 10^{20}

4

2.1 \times 10^{20}

Explanation:

As we know that,
$E = N h ν = N \frac{h c}{λ}$
$N = \frac{E λ}{h c} (1)$
$E = 100 W (J s^{- 1}) \times 1 s = 100 J$
Putting the value in eq.(1) we get
$N = \frac{100 J \times 560 \times 10^{- 9} m}{6.626 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}$
$= 2.82 \times 10^{20}$ photons.

CHXI02:STRUCTURE OF ATOM

307403 Einstein’s photoelectric equation states that $E_{k} = h v - W$ . In this equation, $E_{k}$ refers to

1 Kinetic energy of all ejected electrons

2 Mean kinetic energy of emitted electrons

3 Minimum kinetic energy of emitted electrons

4 Maximum kinetic energy of emitted electrons

CHXI02:STRUCTURE OF ATOM

307404 Assertion :
The energy of quantum of radiation is given by $E = h v$ .
Reason :
Quantum, in the energy equation signifies the principal quantum number.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency $(v)$ and is expressed by the equation. $E = h v$ . So option (3) is correct.

CHXI02:STRUCTURE OF ATOM

307405 The energy of one mole of photons of radiation of wavelength 300nm is
(Given: $h = 6 .63 \times 1 0^{- 34} J s$ , $N_{A} = 6 .02 \times 1 0^{23} m o l^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ )

1

235 k J m o l^{- 1}

2

325 k J m o l^{- 1}

3

399 k J m o l^{- 1}

4

435 k J m o l^{- 1}

Explanation:

Energy of one mole of photons $= \frac{h c}{λ} \times N_{A}$
$= \frac{6 .63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}} \times 6 .02 \times 1 0^{23}$
$= 399 .13 \times 1 0^{3} J o u l e / m o l e$
$= 399 k J / J o u l e$

JEE - 2022

CHXI02:STRUCTURE OF ATOM

307406 The energy required to break 1 mol of Cl-Cl bonds in $C l_{2}$ is $242 k J / m o l .$ The longest wavelength of light capable of breaking a single Cl-Cl bond is

1

594 n m

2

640 n m

3

700 n m

4

494 n m

Explanation:

Energy required for 1 $C l_{2}$ molecule
$= \frac{242 \times 1 0^{3}}{N_{A}} J$
$E = \frac{h c}{λ}$ or $λ = \frac{h c}{E}$
$= \frac{6 .626 \times 1 0^{- 34} \times 3 \times 1 0^{8} \times 6 .02 \times 1 0^{23}}{242 \times 1 0^{3}}$
$= 494 \times 1 0^{- 9} m = 494 n m$

CHXI02:STRUCTURE OF ATOM

307407 Number of photons emitted by a $100 W ({Js}^{- 1})$ yellow lamp in 1.0s is ( $λ$ of yellow light is $560 nm$ )

1

1.6 \times 10^{18}

2

1.4 \times 10^{18}

3

2.8 \times 10^{20}

4

2.1 \times 10^{20}

Explanation:

As we know that,
$E = N h ν = N \frac{h c}{λ}$
$N = \frac{E λ}{h c} (1)$
$E = 100 W (J s^{- 1}) \times 1 s = 100 J$
Putting the value in eq.(1) we get
$N = \frac{100 J \times 560 \times 10^{- 9} m}{6.626 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}$
$= 2.82 \times 10^{20}$ photons.

CHXI02:STRUCTURE OF ATOM

307403 Einstein’s photoelectric equation states that $E_{k} = h v - W$ . In this equation, $E_{k}$ refers to

1 Kinetic energy of all ejected electrons

2 Mean kinetic energy of emitted electrons

3 Minimum kinetic energy of emitted electrons

4 Maximum kinetic energy of emitted electrons

CHXI02:STRUCTURE OF ATOM

307404 Assertion :
The energy of quantum of radiation is given by $E = h v$ .
Reason :
Quantum, in the energy equation signifies the principal quantum number.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency $(v)$ and is expressed by the equation. $E = h v$ . So option (3) is correct.

CHXI02:STRUCTURE OF ATOM

307405 The energy of one mole of photons of radiation of wavelength 300nm is
(Given: $h = 6 .63 \times 1 0^{- 34} J s$ , $N_{A} = 6 .02 \times 1 0^{23} m o l^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ )

1

235 k J m o l^{- 1}

2

325 k J m o l^{- 1}

3

399 k J m o l^{- 1}

4

435 k J m o l^{- 1}

Explanation:

Energy of one mole of photons $= \frac{h c}{λ} \times N_{A}$
$= \frac{6 .63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}} \times 6 .02 \times 1 0^{23}$
$= 399 .13 \times 1 0^{3} J o u l e / m o l e$
$= 399 k J / J o u l e$

JEE - 2022

CHXI02:STRUCTURE OF ATOM

307406 The energy required to break 1 mol of Cl-Cl bonds in $C l_{2}$ is $242 k J / m o l .$ The longest wavelength of light capable of breaking a single Cl-Cl bond is

1

594 n m

2

640 n m

3

700 n m

4

494 n m

Explanation:

Energy required for 1 $C l_{2}$ molecule
$= \frac{242 \times 1 0^{3}}{N_{A}} J$
$E = \frac{h c}{λ}$ or $λ = \frac{h c}{E}$
$= \frac{6 .626 \times 1 0^{- 34} \times 3 \times 1 0^{8} \times 6 .02 \times 1 0^{23}}{242 \times 1 0^{3}}$
$= 494 \times 1 0^{- 9} m = 494 n m$

CHXI02:STRUCTURE OF ATOM

307407 Number of photons emitted by a $100 W ({Js}^{- 1})$ yellow lamp in 1.0s is ( $λ$ of yellow light is $560 nm$ )

1

1.6 \times 10^{18}

2

1.4 \times 10^{18}

3

2.8 \times 10^{20}

4

2.1 \times 10^{20}

Explanation:

As we know that,
$E = N h ν = N \frac{h c}{λ}$
$N = \frac{E λ}{h c} (1)$
$E = 100 W (J s^{- 1}) \times 1 s = 100 J$
Putting the value in eq.(1) we get
$N = \frac{100 J \times 560 \times 10^{- 9} m}{6.626 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}$
$= 2.82 \times 10^{20}$ photons.

CHXI02:STRUCTURE OF ATOM

307403 Einstein’s photoelectric equation states that $E_{k} = h v - W$ . In this equation, $E_{k}$ refers to

1 Kinetic energy of all ejected electrons

2 Mean kinetic energy of emitted electrons

3 Minimum kinetic energy of emitted electrons

4 Maximum kinetic energy of emitted electrons

CHXI02:STRUCTURE OF ATOM

307404 Assertion :
The energy of quantum of radiation is given by $E = h v$ .
Reason :
Quantum, in the energy equation signifies the principal quantum number.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency $(v)$ and is expressed by the equation. $E = h v$ . So option (3) is correct.

CHXI02:STRUCTURE OF ATOM

307405 The energy of one mole of photons of radiation of wavelength 300nm is
(Given: $h = 6 .63 \times 1 0^{- 34} J s$ , $N_{A} = 6 .02 \times 1 0^{23} m o l^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ )

1

235 k J m o l^{- 1}

2

325 k J m o l^{- 1}

3

399 k J m o l^{- 1}

4

435 k J m o l^{- 1}

Explanation:

Energy of one mole of photons $= \frac{h c}{λ} \times N_{A}$
$= \frac{6 .63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}} \times 6 .02 \times 1 0^{23}$
$= 399 .13 \times 1 0^{3} J o u l e / m o l e$
$= 399 k J / J o u l e$

JEE - 2022

CHXI02:STRUCTURE OF ATOM

307406 The energy required to break 1 mol of Cl-Cl bonds in $C l_{2}$ is $242 k J / m o l .$ The longest wavelength of light capable of breaking a single Cl-Cl bond is

1

594 n m

2

640 n m

3

700 n m

4

494 n m

Explanation:

Energy required for 1 $C l_{2}$ molecule
$= \frac{242 \times 1 0^{3}}{N_{A}} J$
$E = \frac{h c}{λ}$ or $λ = \frac{h c}{E}$
$= \frac{6 .626 \times 1 0^{- 34} \times 3 \times 1 0^{8} \times 6 .02 \times 1 0^{23}}{242 \times 1 0^{3}}$
$= 494 \times 1 0^{- 9} m = 494 n m$

CHXI02:STRUCTURE OF ATOM

307407 Number of photons emitted by a $100 W ({Js}^{- 1})$ yellow lamp in 1.0s is ( $λ$ of yellow light is $560 nm$ )

1

1.6 \times 10^{18}

2

1.4 \times 10^{18}

3

2.8 \times 10^{20}

4

2.1 \times 10^{20}

Explanation:

As we know that,
$E = N h ν = N \frac{h c}{λ}$
$N = \frac{E λ}{h c} (1)$
$E = 100 W (J s^{- 1}) \times 1 s = 100 J$
Putting the value in eq.(1) we get
$N = \frac{100 J \times 560 \times 10^{- 9} m}{6.626 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}$
$= 2.82 \times 10^{20}$ photons.

CHXI02:STRUCTURE OF ATOM

307403 Einstein’s photoelectric equation states that $E_{k} = h v - W$ . In this equation, $E_{k}$ refers to

1 Kinetic energy of all ejected electrons

2 Mean kinetic energy of emitted electrons

3 Minimum kinetic energy of emitted electrons

4 Maximum kinetic energy of emitted electrons

CHXI02:STRUCTURE OF ATOM

307404 Assertion :
The energy of quantum of radiation is given by $E = h v$ .
Reason :
Quantum, in the energy equation signifies the principal quantum number.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency $(v)$ and is expressed by the equation. $E = h v$ . So option (3) is correct.

CHXI02:STRUCTURE OF ATOM

307405 The energy of one mole of photons of radiation of wavelength 300nm is
(Given: $h = 6 .63 \times 1 0^{- 34} J s$ , $N_{A} = 6 .02 \times 1 0^{23} m o l^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ )

1

235 k J m o l^{- 1}

2

325 k J m o l^{- 1}

3

399 k J m o l^{- 1}

4

435 k J m o l^{- 1}

Explanation:

Energy of one mole of photons $= \frac{h c}{λ} \times N_{A}$
$= \frac{6 .63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}} \times 6 .02 \times 1 0^{23}$
$= 399 .13 \times 1 0^{3} J o u l e / m o l e$
$= 399 k J / J o u l e$

JEE - 2022

CHXI02:STRUCTURE OF ATOM

307406 The energy required to break 1 mol of Cl-Cl bonds in $C l_{2}$ is $242 k J / m o l .$ The longest wavelength of light capable of breaking a single Cl-Cl bond is

1

594 n m

2

640 n m

3

700 n m

4

494 n m

Explanation:

Energy required for 1 $C l_{2}$ molecule
$= \frac{242 \times 1 0^{3}}{N_{A}} J$
$E = \frac{h c}{λ}$ or $λ = \frac{h c}{E}$
$= \frac{6 .626 \times 1 0^{- 34} \times 3 \times 1 0^{8} \times 6 .02 \times 1 0^{23}}{242 \times 1 0^{3}}$
$= 494 \times 1 0^{- 9} m = 494 n m$

CHXI02:STRUCTURE OF ATOM

307407 Number of photons emitted by a $100 W ({Js}^{- 1})$ yellow lamp in 1.0s is ( $λ$ of yellow light is $560 nm$ )

1

1.6 \times 10^{18}

2

1.4 \times 10^{18}

3

2.8 \times 10^{20}

4

2.1 \times 10^{20}

Explanation:

As we know that,
$E = N h ν = N \frac{h c}{λ}$
$N = \frac{E λ}{h c} (1)$
$E = 100 W (J s^{- 1}) \times 1 s = 100 J$
Putting the value in eq.(1) we get
$N = \frac{100 J \times 560 \times 10^{- 9} m}{6.626 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}$
$= 2.82 \times 10^{20}$ photons.

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