In the photoelectric effect,if the work funct | vedclass

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(B) The work function is the minimum energy required to remove an electron from the surface of a metal. As the atomic size increases down a group,the

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$2.20$

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(B) The work function is the minimum energy required to remove an electron from the surface of a metal. As the atomic size increases down a group,the ionization energy decreases,meaning the work function also decreases.$Li$,$Na$,$K$,and $Rb$ belong to Group $1$ of the periodic table.The order of their atomic sizes is $Li Therefore,the order of their work functions is $Li > Na > K > Rb$.Given values: $Li = 2.41 \ eV$,$Na = 2.30 \ eV$,and $Rb = 2.09 \ eV$.The work function of $K$ must lie between $Na$ $(2.30 \ eV)$ and $Rb$ $(2.09 \ eV)$.Among the given options,$2.20 \ eV$ is the only value that falls within the range $2.09 \ eV < K < 2.30 \ eV$.

Element	$IE_1 + IE_2$ $(kJ/mol)$	$IE_3 + IE_4$ $(kJ/mol)$
$P$	$2.45$	$8.82$
$Q$	$2.85$	$6.11$

In the photoelectric effect,if the work functions of $Li$,$Na$,and $Rb$ are $2.41 \ eV$,$2.30 \ eV$,and $2.09 \ eV$ respectively,the work function of $K$ could approximately be in $eV$?

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The sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:

Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$

$P$ $2.45$ $8.82$

$Q$ $2.85$ $6.11$

Then,according to the given information,the incorrect statement$(s)$ is/are:

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