What is the energy required to move the electron from ground state of H atom to the first excited state? Given that the ground state energy of H atom is 13.6 eV and that the energy En of an electron in n th orbital of an atom or ion of atomic number Z is given by the equation En=-(13.6 Z2 /n2).

Question

What is the energy required to move the electron from ground state of H atom to the first excited state? Given that the ground state energy of H atom is 13.6 eV and that the energy En of an electron in n th orbital of an atom or ion of atomic number Z is given by the equation En=-(13.6 Z2 /n2). (A) 13.6 eV (B) 3.4 eV (C) 10.2 eV (D) -10.2 eV

Tardigrade Admin · Accepted Answer

En=-(13.6/n2) Z2 The energy required to move the electron from ground state (n=1) of H to first excited state (n=2) is Δ E =E2-E1=-13.6((1/22)-(1/12)) =13.6 × (3/4)=10.2 eV

$13.6 e V$

$3.4 e V$

$10.2 e V$

$- 10.2 e V$

$E_{n} = - \frac{13.6}{n ^{2}} Z^{2}$

The energy required to move the electron from ground state $(n = 1)$ of $H$ to first excited state $(n = 2)$ is

$Δ E = E_{2} - E_{1} = - 13.6 (\frac{1}{2 ^{2}} - \frac{1}{1 ^{2}})$

$= 13.6 \times \frac{3}{4} = 10.2$ eV

13.6eV

3.4eV

10.2eV

−10.2eV

En​=−n213.6​Z2 The energy required to move the electron from ground state (n=1) of H to first excited state (n=2) is ΔE=E2​−E1​=−13.6(221​−121​) =13.6×43​=10.2 eV

$13.6 e V$

$3.4 e V$

$10.2 e V$

$- 10.2 e V$

$E_{n} = - \frac{13.6}{n ^{2}} Z^{2}$

The energy required to move the electron from ground state $(n = 1)$ of $H$ to first excited state $(n = 2)$ is

$Δ E = E_{2} - E_{1} = - 13.6 (\frac{1}{2 ^{2}} - \frac{1}{1 ^{2}})$

$= 13.6 \times \frac{3}{4} = 10.2$ eV