Question

When Lithium vapour is taken in a discharge
tube and the potential difference between the
electrodes is 5.4 ev, there is a sudden increase
in the flow of current. The ionisation energy
of Lithium is
1) 54 ev
2) 520 kJ mol-1
3) 54 kJ atom-1 4) 5.4 ev atom-1​

Answer 1

Answer:

1ev=1.6×10^-19 joules

5.4ev=5.4×1.6×10^-19joules

1 atom =6.023×10^23 molecules

then 5.4ev=5.4×10^-19×6.023×10^23jolue per atom

=0.52joule or 520kilojoule

hope it helps u... then mark me as BRAINLIEST

Answer 2

Answer:

520 kJ mol-1

Explanation:

GIVEN- When Lithium vapour is taken in a discharge tube and the potential difference between the electrodes is 5.4 ev, there is a sudden increase in the flow of current.

TO FIND- The ionisation energy

of Lithium

SOLUTION -

1e v=1.6×10^{∧ }-19 joules

5.4ev=5.4×1.6×10^{∧ }-1 9joules

1 atom =6.023×10^{ molecules .4ev=5.4×10^{∧ }-19×6.023×10^{ then

5 jolueper atom=0.52joule or 520kilojoule`

FINAL ANSWER - 520kJ mol-1

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