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CHXI02:STRUCTURE OF ATOM

307377 A light whose frequency is equal to $6 \times 1 0^{14} H z$ is incident on a metal whose work function is 2 eV $(h = 6 .63 \times 1 0^{- 34} J s, 1 e V = 1 .6 \times 1 0^{- 19} J)$ . The maximum energy of electrons emitted will be

1

2 .49 e V

2

4 .49 e V

3

0 .49 e V

4

5 .49 e V

Explanation:

Absorbed energy $(h v) = h v_{o} + K . E .$ of photo-electrons
$\frac{6 .63 \times 1 0^{- 19} \times 6 \times 1 0^{24}}{1 .6 \times 1 0^{- 19}} = 2 e V + K . E .$
$2 .49 e V = 2 e V + K . E .$
$K . E . = 0 .49 e V$

CHXI02:STRUCTURE OF ATOM

307378 With regard to the photoelectric effect, identify the correct statement among the following

1 Numbers of

e^{-}

ejected increases with the increase in the frequency of the incident light

2 Number of

e^{-}

electrons ejected increases with the increase in work function

3 Number of

e^{-}

ejected increases with the increase in the intensity of incident light

4 Energy of

e^{-}

ejected increases with the increase in the intensity of incident light

KCET - 2020

CHXI02:STRUCTURE OF ATOM

307379 Photons of frequency $3.2 \times 10^{16} Hz$ is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is $\frac{3^{th}}{4}$ of the energy of the irradiating photon. What is the threshold frequency of the metal?

1

2.4 \times 10^{25} Hz

2

2.4 \times 10^{16} Hz

3

1.6 \times 10^{15} Hz

4

8 \times 10^{15} Hz

Explanation:

Acc. to photoelectric effect, $E = W + KE$
$∴ W = E - KE$
K.E is $\frac{3}{4}$ of energy supplied
So, $W = \frac{1}{4} E = \frac{1}{4} \times 3.2 \times 10^{16}$
$= 0.8 \times 10^{16}$
$= 8 \times 10^{15} Hz$

CHXI02:STRUCTURE OF ATOM

307380 A metal surface is exposed to solar radiations

1 The emitted electrons has energy less than a maximum value of energy depending upon the frequency of the incident radiation.

2 The emitted electrons have energy less than the maximum value of energy depending upon intensity of incident radiations.

3 The emitted electrons have zero energy.

4 The emitted electrons have energy equal to energy of photons of incident light.

Explanation:

Some amount of given energy is utilised to remove electron. Remaining energy is converted into kinetic energy.
$∴$ Energy of emitted electrons $= E - W$ (work function).

CHXI02:STRUCTURE OF ATOM

307377 A light whose frequency is equal to $6 \times 1 0^{14} H z$ is incident on a metal whose work function is 2 eV $(h = 6 .63 \times 1 0^{- 34} J s, 1 e V = 1 .6 \times 1 0^{- 19} J)$ . The maximum energy of electrons emitted will be

1

2 .49 e V

2

4 .49 e V

3

0 .49 e V

4

5 .49 e V

Explanation:

Absorbed energy $(h v) = h v_{o} + K . E .$ of photo-electrons
$\frac{6 .63 \times 1 0^{- 19} \times 6 \times 1 0^{24}}{1 .6 \times 1 0^{- 19}} = 2 e V + K . E .$
$2 .49 e V = 2 e V + K . E .$
$K . E . = 0 .49 e V$

CHXI02:STRUCTURE OF ATOM

307378 With regard to the photoelectric effect, identify the correct statement among the following

1 Numbers of

e^{-}

ejected increases with the increase in the frequency of the incident light

2 Number of

e^{-}

electrons ejected increases with the increase in work function

3 Number of

e^{-}

ejected increases with the increase in the intensity of incident light

4 Energy of

e^{-}

ejected increases with the increase in the intensity of incident light

KCET - 2020

CHXI02:STRUCTURE OF ATOM

307379 Photons of frequency $3.2 \times 10^{16} Hz$ is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is $\frac{3^{th}}{4}$ of the energy of the irradiating photon. What is the threshold frequency of the metal?

1

2.4 \times 10^{25} Hz

2

2.4 \times 10^{16} Hz

3

1.6 \times 10^{15} Hz

4

8 \times 10^{15} Hz

Explanation:

Acc. to photoelectric effect, $E = W + KE$
$∴ W = E - KE$
K.E is $\frac{3}{4}$ of energy supplied
So, $W = \frac{1}{4} E = \frac{1}{4} \times 3.2 \times 10^{16}$
$= 0.8 \times 10^{16}$
$= 8 \times 10^{15} Hz$

CHXI02:STRUCTURE OF ATOM

307380 A metal surface is exposed to solar radiations

1 The emitted electrons has energy less than a maximum value of energy depending upon the frequency of the incident radiation.

2 The emitted electrons have energy less than the maximum value of energy depending upon intensity of incident radiations.

3 The emitted electrons have zero energy.

4 The emitted electrons have energy equal to energy of photons of incident light.

Explanation:

Some amount of given energy is utilised to remove electron. Remaining energy is converted into kinetic energy.
$∴$ Energy of emitted electrons $= E - W$ (work function).

CHXI02:STRUCTURE OF ATOM

307377 A light whose frequency is equal to $6 \times 1 0^{14} H z$ is incident on a metal whose work function is 2 eV $(h = 6 .63 \times 1 0^{- 34} J s, 1 e V = 1 .6 \times 1 0^{- 19} J)$ . The maximum energy of electrons emitted will be

1

2 .49 e V

2

4 .49 e V

3

0 .49 e V

4

5 .49 e V

Explanation:

Absorbed energy $(h v) = h v_{o} + K . E .$ of photo-electrons
$\frac{6 .63 \times 1 0^{- 19} \times 6 \times 1 0^{24}}{1 .6 \times 1 0^{- 19}} = 2 e V + K . E .$
$2 .49 e V = 2 e V + K . E .$
$K . E . = 0 .49 e V$

CHXI02:STRUCTURE OF ATOM

307378 With regard to the photoelectric effect, identify the correct statement among the following

1 Numbers of

e^{-}

ejected increases with the increase in the frequency of the incident light

2 Number of

e^{-}

electrons ejected increases with the increase in work function

3 Number of

e^{-}

ejected increases with the increase in the intensity of incident light

4 Energy of

e^{-}

ejected increases with the increase in the intensity of incident light

KCET - 2020

CHXI02:STRUCTURE OF ATOM

307379 Photons of frequency $3.2 \times 10^{16} Hz$ is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is $\frac{3^{th}}{4}$ of the energy of the irradiating photon. What is the threshold frequency of the metal?

1

2.4 \times 10^{25} Hz

2

2.4 \times 10^{16} Hz

3

1.6 \times 10^{15} Hz

4

8 \times 10^{15} Hz

Explanation:

Acc. to photoelectric effect, $E = W + KE$
$∴ W = E - KE$
K.E is $\frac{3}{4}$ of energy supplied
So, $W = \frac{1}{4} E = \frac{1}{4} \times 3.2 \times 10^{16}$
$= 0.8 \times 10^{16}$
$= 8 \times 10^{15} Hz$

CHXI02:STRUCTURE OF ATOM

307380 A metal surface is exposed to solar radiations

1 The emitted electrons has energy less than a maximum value of energy depending upon the frequency of the incident radiation.

2 The emitted electrons have energy less than the maximum value of energy depending upon intensity of incident radiations.

3 The emitted electrons have zero energy.

4 The emitted electrons have energy equal to energy of photons of incident light.

Explanation:

Some amount of given energy is utilised to remove electron. Remaining energy is converted into kinetic energy.
$∴$ Energy of emitted electrons $= E - W$ (work function).

CHXI02:STRUCTURE OF ATOM

307377 A light whose frequency is equal to $6 \times 1 0^{14} H z$ is incident on a metal whose work function is 2 eV $(h = 6 .63 \times 1 0^{- 34} J s, 1 e V = 1 .6 \times 1 0^{- 19} J)$ . The maximum energy of electrons emitted will be

1

2 .49 e V

2

4 .49 e V

3

0 .49 e V

4

5 .49 e V

Explanation:

Absorbed energy $(h v) = h v_{o} + K . E .$ of photo-electrons
$\frac{6 .63 \times 1 0^{- 19} \times 6 \times 1 0^{24}}{1 .6 \times 1 0^{- 19}} = 2 e V + K . E .$
$2 .49 e V = 2 e V + K . E .$
$K . E . = 0 .49 e V$

CHXI02:STRUCTURE OF ATOM

307378 With regard to the photoelectric effect, identify the correct statement among the following

1 Numbers of

e^{-}

ejected increases with the increase in the frequency of the incident light

2 Number of

e^{-}

electrons ejected increases with the increase in work function

3 Number of

e^{-}

ejected increases with the increase in the intensity of incident light

4 Energy of

e^{-}

ejected increases with the increase in the intensity of incident light

KCET - 2020

CHXI02:STRUCTURE OF ATOM

307379 Photons of frequency $3.2 \times 10^{16} Hz$ is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is $\frac{3^{th}}{4}$ of the energy of the irradiating photon. What is the threshold frequency of the metal?

1

2.4 \times 10^{25} Hz

2

2.4 \times 10^{16} Hz

3

1.6 \times 10^{15} Hz

4

8 \times 10^{15} Hz

Explanation:

Acc. to photoelectric effect, $E = W + KE$
$∴ W = E - KE$
K.E is $\frac{3}{4}$ of energy supplied
So, $W = \frac{1}{4} E = \frac{1}{4} \times 3.2 \times 10^{16}$
$= 0.8 \times 10^{16}$
$= 8 \times 10^{15} Hz$

CHXI02:STRUCTURE OF ATOM

307380 A metal surface is exposed to solar radiations

1 The emitted electrons has energy less than a maximum value of energy depending upon the frequency of the incident radiation.

2 The emitted electrons have energy less than the maximum value of energy depending upon intensity of incident radiations.

3 The emitted electrons have zero energy.

4 The emitted electrons have energy equal to energy of photons of incident light.

Explanation:

Some amount of given energy is utilised to remove electron. Remaining energy is converted into kinetic energy.
$∴$ Energy of emitted electrons $= E - W$ (work function).

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