When photon of energy 4. 0eV strikes the surface of a metal A , the...

From Einstein photoelectric equation,
E = {ϕ}_{0} + {KE}_\max

For metal $A4 = {ϕ}_{A} + {T}_{A}.... \left(i\right)$
For metal $B4. 5 = {ϕ}_{B} + \left({T}_{A} -1. 5\right).... \left(ii\right)$

From Eq. (i) and (ii), wo got
${ϕ}_{B} - {ϕ}_{A} = 2$

Now, according to de-Broglie hypothesis,
${λ}_{A} = \dfrac{h}{mv} = \dfrac{h}{2m{T}_{A}}$

Similarly, ${λ}_{B} = \dfrac{h}{2m{T}_{B}}$

∴ \dfrac{\lambdaA}{\lambdaB} = \sqrt{\dfrac{TB}{TA}} = \sqrt{\dfrac{{T}_{A} - 1. 5}{TA}} = \left(1 -\dfrac{1. 5}{TA}\right)1 / 2$</p><p> \left(\dfrac{1}{2}\right)$2 = 1 - $\dfrac{1. 5}{TA}$$</p><p>On solving,${T}_{A} = 2. 0eV$</p><p>or${ϕ}_{A} = 4 - {T}_{A} = 4 - 2 = 2. 0eV$</p><p>${ϕ}_{B} = 6 - {T}_{A} = 6 - 2 = 4. 0eV