The photoelectric work function of a metal is 3.3 eV. The threshold frequency for this metal is approximately

Question

The photoelectric work function of a metal is 3.3 eV. The threshold frequency for this metal is approximately (A)  3.3× 1013 Hz  (B)  8.0× 1014 Hz  (C)  1.65× 1015 Hz  (D)  9.9× 1015 Hz

Tardigrade Admin · Accepted Answer

Given, f=3.3eV=3.3× 1.6× 10-19V We know that f=hv v=(f/h) v=(3.3× 1.6× 10-19/6.6× 10-34)=0.8× 1015Hz v=8.0× 1014Hz

Q. The photoelectric work function of a metal is 3.3 eV. The threshold frequency for this metal is approximately

$3.3 \times 10^{13} Hz$

$8.0 \times 10^{14} Hz$

$1.65 \times 10^{15} Hz$

$9.9 \times 10^{15} Hz$

Given, $f = 3.3 e V = 3.3 \times 1.6 \times 10^{- 19} V$ We know that $f = h v$ $v = \frac{f}{h}$ $v = \frac{3.3 \times 1.6 \times 10 ^{- 19}}{6.6 \times 10 ^{- 34}} = 0.8 \times 10^{15} Hz$ $v = 8.0 \times 10^{14} Hz$

Q. The photoelectric work function of a metal is 3.3 eV. The threshold frequency for this metal is approximately

3.3×1013Hz

8.0×1014Hz

1.65×1015Hz

9.9×1015Hz

Given, f=3.3eV=3.3×1.6×10−19V We know that f=hv v=hf​ v=6.6×10−343.3×1.6×10−19​=0.8×1015Hz v=8.0×1014Hz

$3.3 \times 10^{13} Hz$

$8.0 \times 10^{14} Hz$

$1.65 \times 10^{15} Hz$

$9.9 \times 10^{15} Hz$

Given, $f = 3.3 e V = 3.3 \times 1.6 \times 10^{- 19} V$ We know that $f = h v$ $v = \frac{f}{h}$ $v = \frac{3.3 \times 1.6 \times 10 ^{- 19}}{6.6 \times 10 ^{- 34}} = 0.8 \times 10^{15} Hz$ $v = 8.0 \times 10^{14} Hz$