KVPY 2017 Chemistry Question Paper with Answer and Solution

(B) The energy of the incident photon is calculated as $E = \frac{hc}{\lambda}$.
Given $\lambda = 400 \ nm = 400 \times 10^{-9} \ m$.
$E = \frac{6.626 \times 10^{-34} \ J \ s \times 3 \times 10^8 \ m \ s^{-1}}{400 \times 10^{-9} \ m} = 4.97 \times 10^{-19} \ J$.
Converting to $eV$: $E = \frac{4.97 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ J \ eV^{-1}} \approx 3.1 \ eV$.
Photoelectrons are emitted if the incident photon energy is greater than or equal to the work function of the metal.
Based on typical work functions ($K \approx 2.2 \ eV$,$Li \approx 2.4 \ eV$,$Mg \approx 3.7 \ eV$,$Ag \approx 4.3 \ eV$),only $K$ and $Li$ have work functions less than $3.1 \ eV$.

(C) The heat lost by the metal is equal to the heat gained by the water in an insulated container.
Let $c$ be the specific heat of the metal.
Mass of metal $m_m = 100 \, g$,Initial temperature $T_{m,i} = 80^{\circ} C$.
Mass of water $m_w = 1000 \, g$,Initial temperature $T_{w,i} = 15^{\circ} C$.
Final equilibrium temperature $T_f = 15.69^{\circ} C$.
Specific heat of water $c_w = 4.184 \, J / g \cdot ^{\circ} C$.
Heat lost by metal $= m_m \times c \times (T_{m,i} - T_f) = 100 \times c \times (80 - 15.69) = 6431c$.
Heat gained by water $= m_w \times c_w \times (T_f - T_{w,i}) = 1000 \times 4.184 \times (15.69 - 15) = 4184 \times 0.69 = 2886.96 \, J$.
Equating the two: $6431c = 2886.96$.
$c = \frac{2886.96}{6431} \approx 0.4489 \, J / g \cdot ^{\circ} C$.
Rounding to two decimal places,we get $0.45 \, J / g \cdot ^{\circ} C$.