Question

the electron in a given bohr orbit has a total energy of - 1.5ev calculate its kinetic energy and potential energy and wavelength of radiation emitted when this electron makes a transition to ground state

Answer 1

here is your ans.............

.

Answer 2

Question:

The electron in a given Bohr orbit has a total energy of $-1.5 \mathrm{eV}$. Calculate its (i) Kinetic energy.

(ii) potential energy.

(iii) Wavelength of radiation emitted, when this electron makes a transition to the ground state. [Given : Energy in the ground state $=-13.6 \mathrm{eV}$ and Rydberg's constant $=1.09 \times 10^{7} \mathrm{~m}^{1}$ ]

Answer:

(i) Kinetic energy of the electron = 1.5 ev.

(ii) The potential energy of the electron = -3.0 eV.

(iii) Wavelength of гadiation $\lambda=1.022 \times 10^{-7} \mathrm{~m}$.

Explanation:

Given:

E = 1.5 ev

(i) Kinetic energy of the electron

Kinetic energy = -E

Kinetic energy = -5(-1.5 ev)

Kinetic energy = 1.5 ev

(ii) The potential energy of the electron

Potential energy = -2 K.E.

Potential energy = -2 $\times$ 1.5

Potential energy = -3.0 eV

(iii) Energy of photon

$\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}$

$&\Delta \mathrm{E}=-1.5-(-13.6)$

$&\Delta \mathrm{E}=12.1 \mathrm{ev}$

$&\Delta \mathrm{E}=12.1 \times 1.6 \times 10^{-19} \mathrm{~J}$

$&\Delta \mathrm{E}=19.36 \times 10^{-19} \mathrm{~J}$

Energy of photon

$&\Delta \mathrm{E}=\frac{h c}{\lambda}$

$&\frac{h c}{\lambda}=19.36 \times 10^{-19}$

Wavelength of гadiation

$\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{19.36 \times 10^{-19}}$

$\lambda=1.022 \times 10^{-7} \mathrm{~m}$.

Therefore,

(i) Kinetic energy of the electron = 1.5 ev.

(ii) The potential energy of the electron = -3.0 eV.

(iii) Wavelength of гadiation $\lambda=1.022 \times 10^{-7} \mathrm{~m}$.

SPJ2