The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (given ionization energy of $H = 2.18 \times 10^{- 18} J a t o m^{1}$ and $h = 6.625 \times 10^{- 34} J s$

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

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

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

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

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Solution

The correct option is C $3.08 \times 10^{15} s^{- 1}$
Hence, Option "C" is the correct answer.

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Similar questions

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be

(Given ionization energy of $H = 2.18 \times 10^{- 18} J a t o m^{- 1}$ and $h = 6.625 \times 10^{- 34} J s$ )

Q. The frequency of radiation emitted when the electron falls from

n = 4

n = 1

in a hydrogen atom will be: (Given ionization energy of

H = 2.18 \times 10^{- 18} J a t o m^{- 1}

and

h = 6.625 \times 10^{- 34} J s)

Q. The frequency of radiation emitted when the electron falls from

n = 4

n =

lina hydrogen atom will be (Given ionisation energy of

H = 2.18 10^{- 18} J a t o m^{- 1}

and

h = 6.625 \times 10^{- 34} J s)

Q. Calculate the minimum frequency required to remove a first shell electron completely from a hydrogen atom. Given: R

= 1.097 \times 10^{- 2}

^{- 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

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The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom will be (given ionization energy of H=2.18×10−18J atom1 and h=6.625×10−34 J s

The correct option is C 3.08×1015s−1Hence, Option "C" is the correct answer.

The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (given ionization energy of $H = 2.18 \times 10^{- 18} J a t o m^{1}$ and $h = 6.625 \times 10^{- 34} J s$

The correct option is C $3.08 \times 10^{15} s^{- 1}$
Hence, Option "C" is the correct answer.