Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = $2$ to n = $1$ transition has energy $74.8$ eV higher than the photon emitted in the n = $3$ to n = $2$ transition. The ionization energy of the hydrogen atom is $13.6$ eV. The value of Z is ______.

Open in App

Solution

$△ E_{2 - 1} = 13.6 \times z^{2} [1 - \frac{1}{4}] = 13.6 \times z^{2} [\frac{3}{4}]$

$△ E_{3 - 2} = 13.6 \times z^{2} [\frac{1}{4} - \frac{1}{9}] = 13.6 \times z^{2} [\frac{5}{36}]$

$△ E_{2 - 1} = △ E_{3 - 2} + 74.8$

$13.6 \times z^{2} [\frac{3}{4}] = 13.6 \times z^{2} [\frac{5}{36}] + 74.8$

$13.6 \times z^{2} [\frac{3}{4} - \frac{5}{36}] = 74.8$

$z^{2} = 9$

$z = + 3$ (z can't be - ve)

Suggest Corrections

Similar questions

Q. A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. It makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Then n and Z for this atom will be (Ground state energy of hydrogen atom is – 13.6 eV)

Q. A hydrogen-like atom of atomic number

Z

is in an excited state of quantum number

2 n

. It can emit a maximum energy photon of

204 e V

. If it makes a transition to quantum state

n

, a photon of energy

40.8 e V

is emitted. Find

n, Z

and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is

- 13.6 e V

Q. What is the energy, momentum and wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from

n = 2

n = 1

? Given that ionization potential is

13.6 e V

A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2 n$ . It can emit a photon of maximum energy 204 $e V$ . If it makes a transition to quantum state $n$ , a photon of energy 40.8 $e V$ is emitted. The value of $n$ will be.

Q. A Hydrogen like atom of atomic number Z is in an excited state 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Identity correct statement(s).

Join BYJU'S Learning Program

△E2−1=13.6×z2[1−14]=13.6×z2[34] △E3−2=13.6×z2[14−19]=13.6×z2[536] △E2−1=△E3−2+74.8 13.6×z2[34]=13.6×z2[536]+74.8 13.6×z2[34−536]=74.8 z2=9 z=+3 (z can't be - ve)