Consider a metal that has a work function of 1.2 eV. What is the maximum kinetic

Question

Consider a metal that has a work function of 1.2 eV. What is the maximum kinetic energy (in eV) of photoelectrons that are emitted by this metal, when photons with a wavelength of 400 nm shine on this metal? (a) 3.1 (b) -1.9 (c) 1.9 (d) 0 (e) none of the above An electron (with mass equal to 9.11 times 10^-33 kg) has a speed of 0.5 e. Determine the difference between its relativistic kinetic energy and its non-relativistic kinetic energy, in Joules. (a) 3.0 times 10^-15 (b) 2.0 times 10^-15 (c) 1.5 times 10^-15 (d) 1.8 times 10^-15 (e) 2.4 times 10^-15 The most common isotope of is Its, which has a half life of 3.82 days. What fraction of these original radioactive have still not decayed after one week? (a) 0.281 (b) 0.419 (c) 0.181 (d) 0.834 (e) 0.761 In SI the electric field is an electromagnetic wave in empty space is described by E_ = 100 sin (1.0 times 10^7 z - wt). What is the frequency, f, of this wave (is Hertx)? (a) 100 (b) .01 (c) 1.0 times 10^7 (d) 6.28 times 10^-7 (e) 4.78 times 10^14 Suppose a person stands 5 m in front of a concave mirror. This mirror has a radius of curvature of 100 cm. How tall will this person appear in the mirror (cm), if this person is 2 m tall? (a) 68 (b) 48 (c) 18 (d) 88 (e) 22

Explanation / Answer

Q6.

energy of photon of wavelength 400 nm=h*c/wavelength

where h=planck's constant

c=speed of light

energy=3.106 eV

so maximum energy=energy of incident light-work function

=1.906 eV

hence option c is correct.

Q7.

lorentz' factor=1/sqrt(1-0.5^2)=1.1547

relativistic kinetic energy=(1.1547-1)*mass*c^2

non-relativistic kinetic energy=0.5*mass*speed^2

=0.5*mass*0.5^2*c^2

=0.125*mass*c^2

difference=(0.1547-0.125)*mass*c^2

=2.4351*10^(-15) J

option e is correct.

Q8.

number of nuclei undecayed=A*(exp(-lambda*t))

where lambda=decay constant=ln(2)/half life

so for t=1 week,

number of nuclei undecayed=0.281*A

so option a is correct.

Q9.

wavelength=2*pi/(10^7)

speed=3*10^8 m/s

then frequency=speed/wavelegngth=4.78*10^14 Hz

option e is correct.

Q10.

object distance=u=-5 m

focal length=f=-100 cm/2=-0.5 m

if image distance is v,

then using mirror equation:

(1/v)+(1/u)=1/f

==>v=-0.45445 m

so magnification magnitude=v/u=0.09

so image height=magnification*object height

=0.09*200 cm=18 cm

otpion c is correct.

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