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Q.

The work function of cesium is $2.14 e V .$ The threshold frequency of cesium is:

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a

$5.16 \times 10^{19} Hz$

b

$5.16 \times 10^{16} Hz$

c

$5.16 \times 10^{14} Hz$

d

$5.16 \times 10^{18} Hz$

answer is D.

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Detailed Solution

Here, $ϕ_{0} = 2.14 e V$

$ν_{0} = \frac{ϕ_{0}}{h} = \frac{2.14 \times 1.6 \times 10^{- 19}}{6.63 \times 10^{- 34}} = 5.16 \times 10^{14} Hz$

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