Question

when light of frequency 3.2 into 10 to the power 16 Hz is used to irradiate a metal surface .the max kinetic energy of the emitted photoelectron in 3/4 of the energy of the irradiating photon .what is the threshold frequency of the metal ?​

Answer 1

Answer:

$8 \times {10}^{5} hz$

Explanation:

as you have learn

KE= hv-hv°

where v is frequency given to emitted

and v° is your threshold frequency

also it given that

KEmax= 3/4 E =0.75E

where E = hv =

$6.62 \times {10}^{ - 34} \times 3.2 \times {10}^{16} \\ = 21.184 \times {10}^{ - 18}$

now KE= hv- hv°

so hv°= hv-KE

= E-0.75E= 0.25×

$21.184 \times {10}^{ - 18}$

so hv°=

$5.296 \times {10}^{ - 18}$

now v°=

$\frac{5.296 \times {10}^{ - 18} }{6.62 \times {10}^{ - 34} } = 8 \times {10}^{5} hz$

answer

Answer 2

Answer:

for frequency= 1.6*10^16s-¹

E1= h υ

= h*1.6*10^16s-¹

KE= hυ+hυo

KE1=h*1.6*10^16s-¹+hυo

for frequency= 1.0"10^16s-¹

E2= h*1.0*10^16s-¹

KE2=h*1.0*10^16s-¹+hυo

according to question

KE1=2KE2

h*1.6*10^16s-¹+hυo=2(h*1.0*10^16s-¹+hυo)

h*1.6*10^16s-¹+hυo=2*h*1.0*10^16s-¹+2hυo

h(1.6-1.0)*10^16=hυo

υo=0.6*10^16s-¹

= 6*10^15s-¹

Explanation: