The maximum kinetic energy of the photoelectrons is found to be $6.63 \times 10^{- 19} J$ , when the metal is irradiated with a radiation of frequency $2 \times 10^{15} H z$ . The threshold frequency of the metal is about:

1 \times 10^{15} s^{- 1}

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2 \times 10^{15} s^{- 1}

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3 \times 10^{15} s^{- 1}

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1.5 \times 10^{15} s^{- 1}

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Solution

The correct option is A $1 \times 10^{15} s^{- 1}$
$K E = h (v - v_{0})$
$v_{0} = v - \frac{K E}{h} = 2 \times 10^{15} - \frac{6.63 \times 10^{- 19}}{6.63 \times 10^{- 34}} = 1 \times 10^{15} s^{- 1}$ .

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Similar questions

The maximum kinetic energy of the photoelectrons is found to be $6.63 \times 10^{- 19} J$ . When the metal is irradiated with a radiation of frequency $2 \times 10^{15} H z$ , the threshold frequency of the metal is about

Q. The maximum kinetic energy of the photoelectrons is found to be

6.63 \times 10^{- 19} J

, when the metal is irradiated with a radiation of frequency

2 \times 10^{15} H z

The threshold frequency of the metal is about

Q. A metal was irradiated by light of frequency

3.25 \times 10^{15} s^{- 1}

. The photoelectron produced had its kinetic energy, 2 times the kinetic energy of the photoelectrons which was produced when the same metal was irradiated with a light of frequency

2 \times 10^{15} s^{- 1}

. What is its work function
[Given:

N_{A} = 6 \times 10^{23}, h = 6.6 \times 10^{- 34} J s

]

Q.
The threshold frequency of a metal is

6 \times 10^{14} s^{- 1}

. Calculate the kinetic energy of an electron emitted when radiation of frequency

1.1 \times 10^{15} s^{- 1}

hits the metal.

The threshold frequency $ν_{0}$ for a metal is $7.0 \times 10^{14} s^{- 1}$ . Calculate the kinetic energy of an electron emitted when radiation of frequency $ν$ = $1.0 \times 10^{15} s^{- 1}$

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The maximum kinetic energy of the photoelectrons is found to be 6.63×10−19J, when the metal is irradiated with a radiation of frequency 2×1015Hz. The threshold frequency of the metal is about:

The correct option is A 1×1015s−1KE=h(v−v0)v0=v−KEh=2×1015−6.63×10−196.63×10−34=1×1015s−1.

The maximum kinetic energy of the photoelectrons is found to be $6.63 \times 10^{- 19} J$ , when the metal is irradiated with a radiation of frequency $2 \times 10^{15} H z$ . The threshold frequency of the metal is about:

The correct option is A $1 \times 10^{15} s^{- 1}$
$K E = h (v - v_{0})$
$v_{0} = v - \frac{K E}{h} = 2 \times 10^{15} - \frac{6.63 \times 10^{- 19}}{6.63 \times 10^{- 34}} = 1 \times 10^{15} s^{- 1}$ .