A photon collides with a stationary hydrogen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \, eV$. The first excited state is $E_2 = -3.4 \, eV$. The energy required to exci

Vedclass · Accepted Answer

One photon of energy $10.2 \, eV$ and an electron of energy $1.4 \, eV$

Vedclass · Answer

(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \, eV$. The first excited state is $E_2 = -3.4 \, eV$. The energy required to excite the atom from ground state to the first excited state is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.$2$. When the first photon of $10.2 \, eV$ collides,the atom absorbs it and transitions to the first excited state. After a microsecond,the atom de-excites,emitting a photon of $10.2 \, eV$.$3$. When the second photon of $15 \, eV$ collides with the atom (which is now back in the ground state),since $15 \, eV > 13.6 \, eV$ (the ionization energy of hydrogen),the atom undergoes ionization.$4$. The kinetic energy of the ejected electron is $K = E_{photon} - |E_1| = 15 \, eV - 13.6 \, eV = 1.4 \, eV$.$5$. Therefore,the detector will observe one photon of $10.2 \, eV$ (from the first event) and one electron of $1.4 \, eV$ (from the second event).

Similar Questions

The wavelengths of electron waves in two orbits is $3: 5$. The ratio of kinetic energy of electrons will be

Excess of which of the following in water causes blue baby syndrome disease?

If $\alpha, \beta, \gamma$ are the roots of $2x^3 - 2x - 1 = 0$,then $(\Sigma \alpha \beta)^2$ is equal to

What does the given structural formula represent?

Resin and turpentine are obtained from

Vedclass Test Series

Exam Paper Generator

Online Exam Module