Finding Velocity of Photo-electron from H-Atom & Li2+ Ion

[Saitama]

Homework Statement

If the energy of H-atom in the ground state is -E, the velocity of the photo-electron emitted when a photon having energy E_P strikes a stationary Li²⁺ ion in the ground state is given by:
(a)$v=\sqrt{\frac{2(E_P-E)}{m}}$
(b)$v=\sqrt{\frac{2(E_P+9E)}{m}}$
(c)$v=\sqrt{\frac{2(E_P-9E)}{m}}$
(d)$v=\sqrt{\frac{2(E_P-3E)}{m}}$

Homework Equations

E=W+K.E.
(W represents work function)

The Attempt at a Solution

Energy of H-atom in ground state, $E_H=-E$
Energy of Li⁺² ion in ground state, $E_Li=-2.18*10^{-18}J*3^2=-9E$

$$K.E.=E-W$$
$$\frac{1}{2}mv^2=E_P-(-9E)$$
$$v=\sqrt{\frac{2(E_P+9E)}{m}}$$

But the answer in the answer key is option (c).

 

[pmsrw3]

The sign of the energy is a little confusing in this problem. Remember, energy is conserved, so the total energy before the photon is absorbed is that same as the total energy after. "Total" means you add all the relevant energies. But remember also that the energy of the atom before absorption of the photon is negative.

I would recommend that, instead of the photoelectric equation, you use the conservation of energy equation.

 

[Saitama]

The sign of the energy is a little confusing in this problem. Remember, energy is conserved, so the total energy before the photon is absorbed is that same as the total energy after. "Total" means you add all the relevant energies. But remember also that the energy of the atom before absorption of the photon is negative.

I would recommend that, instead of the photoelectric equation, you use the conservation of energy equation.

Sorry, i didn't get you.
Which is the conservation of energy equation?

 

[pmsrw3]

Which is the conservation of energy equation?

Total energy before = total energy after

 

[Saitama]

Total energy before = total energy after

I know this but what formulae should i use here?

 

[Saitama]

I have found the mistake in my solution.
-9E is the energy of Li²⁺ ion.
Therefore i need +9E energy to remove an electron.
So the work function becomes +9E rather than -9E and i get the answer as (c) option.

But would you please tell me your way to solve it?

 

[I like Serena]

Hi Pranav-Arora!

The energy before (initial energy E_i) is the energy of the photon plus the energy of the Lithium ion, which is E_i = E_p + (-9E).

The energy after (final energy E_f) is the kinetic energy of the electron plus the new energy of the Lithium ion, which is E_f = (1/2)mv² + 0.

Energy conservation says E_i = E_f.
So E_p + (-9E) = (1/2)mv² + 0.

From this you can deduce v.

 

[Saitama]

Hi Pranav-Arora!

The energy before (initial energy E_i) is the energy of the photon plus the energy of the Lithium ion, which is E_i = E_p + (-9E).

The energy after (final energy E_f) is the kinetic energy of the electron plus the new energy of the Lithium ion, which is E_f = (1/2)mv² + 0.

Energy conservation says E_i = E_f.
So E_p + (-9E) = (1/2)mv² + 0.

From this you can deduce v.

Got it. Thanks for the explanation.