(1) 1.8
(2) 1.4
(3) 2.5
(4) 5.6
Ans: (1) 1.8
Sol: ${hc\over\lambda_1}=\phi+{1\over2}m(2v)^2$
${hc\over\lambda_2}=\phi+{1\over2}mv^2$
$\implies{{hc\over\lambda_1}-\phi\over{hc\over\lambda_2}-\phi}=4$
$\implies{hc\over\lambda_1}-\phi={4hc\over\lambda_2}-4\phi$
$\implies{4hc\over\lambda_2}-{hc\over\lambda_1}=3\phi$
$\implies\phi={1\over3}hc\left({4\over\lambda_2}-{1\over\lambda_1}\right)$
$={1\over3}\times1240\left({4\times350-540\over350\times540}\right)$
$=1.8\text{ eV}$