In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P , Q and R are EP, EQ and ER, respectively, and they are related by EP=2 EQ=2 ER. In this experiment, the same source of monochromatic light is used for metal P and Q while a different source of monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV , 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in e V, is .

Question

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P , Q and R are EP, EQ and ER, respectively, and they are related by EP=2 EQ=2 ER. In this experiment, the same source of monochromatic light is used for metal P and Q while a different source of monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV , 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in e V, is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let E R = E Then E Q = E E P =2 E From Einstein's equation for P , Q and R 2 E = hv -4.0 ldots (i) E = hv -4.5 ldots (ii) E = hv prime-5.5 ...(iii) Solving equation (i) and (ii) we get E =0.5 So, hv prime=6.0 eV

Let ER​=E Then EQ​=E EP​=2E From Einstein's equation for P,Q and R 2E=hv−4.0… (i) E=hv−4.5… (ii) E=hv′−5.5...(iii) Solving equation (i) and (ii) we get E=0.5 So, hv′=6.0eV

Let $E_{R} = E$
Then $E_{Q} = E$
$E_{P} = 2 E$
From Einstein's equation for $P, Q$ and $R$
$2 E = h v - 4.0 \dots$ (i)
$E = h v - 4.5 \dots$ (ii)
$E = h v^{'} - 5.5$ ...(iii)
Solving equation (i) and (ii) we get
$E = 0.5$
So, $h v^{'} = 6.0 e V$