(1) 1.8

(2) 1.4

(3) 2.5

(4) 5.6

(1) 1.8

(2) 1.4

(3) 2.5

(4) 5.6

**Ans: (1)** 1.8

**Sol:** ${hc\over\lambda_1}=\phi+{1\over2}m(2v)^2$

${hc\over\lambda_2}=\phi+{1\over2}mv^2$

$\implies{{hc\over\lambda_1}-\phi\over{hc\over\lambda_2}-\phi}=4$

$\implies{hc\over\lambda_1}-\phi={4hc\over\lambda_2}-4\phi$

$\implies{4hc\over\lambda_2}-{hc\over\lambda_1}=3\phi$

$\implies\phi={1\over3}hc\left({4\over\lambda_2}-{1\over\lambda_1}\right)$

$={1\over3}\times1240\left({4\times350-540\over350\times540}\right)$

$=1.8\text{ eV}$