Question

A certain metal when irradiated by light ($\upsilon = 3.2 \times 10^{6}$Hz). Emits photoelectrons with twice K.E. as did photoelectrons when the same metal is irradiated by light ($\upsilon = 2.0 \times 10^{16}$Hz). The $\upsilon_{o}$of the metal is.

• A. $1.2 \times 10^{14}Hz$
• B. $8 \times 10^{15}Hz$
• C. $1.2 \times 10^{16}Hz$
• D. $4 \times 10^{12}Hz$

Answer 1

Answer: (B)

$8 \times 10^{15}Hz$

Answer 2

: $(K.E.)_{1} = h\upsilon_{1} - h\upsilon_{0}$

$(K.E.)_{2} = h\upsilon_{2} - h\upsilon_{0}$

As $(K.E.)_{1} = 2 \times (K.E.)_{2}$

$\therefore(h\upsilon_{1} - h\upsilon_{0}) = 2(h\upsilon_{2} - h\upsilon_{0})$

or $\upsilon_{0} = 2\upsilon_{2} - \upsilon_{1}$

$= 2 \times (2 \times 10^{16}) - (3.2 \times 10^{16})$

$= 0.8 \times 10^{16}Hzor8 \times 10^{15}Hz$