Lithium, beryllium, and mercury have work functions of 2.30 eV,3.90 eV, and 4.50

Question

Lithium, beryllium, and mercury have work functions of 2.30 eV,3.90 eV, and 4.50 eV, respectively. If 352 nm light is incident on each of these metals,determine the following. (a) which metals exhibit the photoelectriceffect
1 none
lithium only
lithium and beryllium
mercury only
mercury and beryllium
all three

(b) the maximum kinetic energy of the photoelectrons in each case(Enter 0 if not applicable.) Lithium 2 eV Beryllium 3 eV Mercury 4 eV
1 none
lithium only
lithium and beryllium
mercury only
mercury and beryllium
all three 1 none
lithium only
lithium and beryllium
mercury only
mercury and beryllium
all three 1 Lithium 2 eV Beryllium 3 eV Mercury 4 eV

Explanation / Answer

(a) the energy of the photon is   hc/ = 4.14 x 10-15 * 3.0 x108 / 352 x 10-9 =   3.528 eV .    Only Lithium is less than this, so only Lithium exhibits the PEeffect . (b)   you already know   Beryllium and Mercury are  zero . So for Lithium...    max KE = 3.528- 2.30 =    1.228eV

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---