When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA - 1.5)eV . If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is :

Question

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA - 1.5)eV . If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is : (A) 1.5 eV (B) 4 eV (C) 3 eV (D) 2 eV

Tardigrade Admin · Accepted Answer

λB=2λA ⇒ (h/√2TBm)=(2h/√2TAm) TA=4TB ...(i) and TB = (TA- 1.5) eV ...(ii) from (i) and (ii) 3TB 1.5 eV ⇒ TB=0.5 eV TB=0.5 eV=4.5 eV-φB φ=4 eV

1.5 eV

4 eV

3 eV

2 eV

λB​=2λA​ ⇒2TB​m​h​=2TA​m​2h​ TA​=4TB​...(i) and TB​=(TA​−1.5)eV...(ii) from (i) and (ii) 3TB​1.5eV⇒TB​=0.5eV TB​=0.5eV=4.5eV−ϕB​ ϕ=4eV

$λ_{B} = 2 λ_{A}$
$\Rightarrow \frac{h}{2 T _{B} m} = \frac{2 h}{2 T _{A} m}$
$T_{A} = 4 T_{B} ... (i)$
and $T_{B} = (T_{A} - 1.5) e V ... (ii)$
from $(i)$ and $(ii)$
$3 T_{B} 1.5 e V \Rightarrow T_{B} = 0.5 e V$
$T_{B} = 0.5 e V = 4.5 e V - ϕ_{B}$
$ϕ = 4 e V$