Question 1

A Hydrogen-like atom has atomic number Z . Photons emitted in the electronic transitions from level n = 4 to level n = 3 in these atoms are used to perform photoelectric effect experiment on a target

Accepted Answer

From Einstein's photoelectric effect, we can write E = K E m a x + &#981; &#8658; E = K E m a x + h c &#955; &#8658; &#8710; E 4 &#160; to &#160; 3 = 1 . 95 &#160; eV + 1240 310 eV &#8658; &#8710; E 4 &#160; to &#160; 3 = 5 . 95 &#160; eV Now, as we know for Hydrogen like atoms, we can write  &#8710; E = 13 . 6 &#160; Z 2 1 n 1 2 - 1 n 2 2 Therefore, &#8658; 13 . 6 &#160; Z 2 1 3 2 - 1 4 2 = 5 . 95 &#8658; 13 . 6 &#160; Z 2 7 9 &#215; 16 = 5 . 95 &#8658; Z 2 = 5 . 95 &#215; 9 &#215; 16

Question 2

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6 . 0 &#160; V . This potential drops to 0 . 6 &#160; V  if

Accepted Answer

According to the Einstein's equation of photoelectric effect for the first case, h c &#955; - &#981; = 6 &#160; eV &#160; &#160; &#8943; ( i ) And for the second case, h c 4 &#955; - &#981; = 0 . 6 &#160; eV Therefore, 3 h c 4 &#955; = 5 . 4 eV &#8658; &#160; &#955; = 3 h c 4 &#215; 5 . 4 eV = 3 &#215; 1 . 24 &#215; 10 - 6 4 &#215; 5 . 4 &#8658; &#955; = 1 . 72 &#215; 10 - 7 &#160; m &#8658;  from equation  i h c 1 . 72 &#215; 10 - 7 &#215; 1 1 . 6 &#215; 10 - 19 - &#981; = 6 &#160

Question 3

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals  P , Q  and  R  are  E P , &#160; E Q  and  E R , respectively, and they ar

Accepted Answer

Let the maximum kinetic energies of photoelectrons be E P = 2 E Q = 2 E R = 2 E . The energy of incident photon on metal  P &#160; and &#160; Q  be  U  and for  R  be  U ' . Photoelectric equation,  E ph = &#981; + K E max Now, using it for the three cases,  U = 4 + 2 E &#160; &#160; &#160; . . . 1 U = 4 . 5 + E &#160; &#160; &#160; . . . 2 U ' = 5 . 5 + E &#160; &#160; &#160; . . . 3 From  1 &#160; and &#160; 2 ,  E = 0 . 5 &#

Dual Nature of Matter — Previous Year Questions & Solutions