A photon of energy 5.4852 eV liberates an electron from the Li-atom initially at rest. The emitted electron moves at right angles to the direction in which the photon moves. Find the speed and the direction in which the Li 2+ ion will move. Ionization potential of Li-atom 5.3918 V, atomic weight = 6.94 g, N = 6.02 × 10 23 /mol and m e = 9.1 × 10 –31 kg.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Energy of the emitted electron
= 
mv 2 = (5.4852 – 5.3918) × 1.6 × 10 –19 = 0.0934 × 1.6 × 10
–19 J
∴ v = 
= 0.18 × 10 6 m/s
If the ion moves with velocity V at angle θ with the direction of the photon, the photon, then, from
conservation of momentum, MV sin θ = mv
where M is the mass of the lithium ion.
∴ V sin θ = 
On substituting the values, we have
V sin θ = 
× 6.02 × 10 23 = 
× 6.02 = 14.2 m/s
Similarly, MV cos θ = 
= 
∴ V cos θ = 
= 0.2538 m/s
V = 
= 14.2 m/s
tan θ = V sin θ /V cos θ = 55.95
θ = 89º
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