Question

The threshold frequency(mu not) for a metal is 7.0 x10s.
Calculate the kinetic energy
emitted when radiation of frequency 1.0×10^15
hits this metal.

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Answer 1

Answer:

Given that, 

vo=7.0×1014s−1

v=1.0×1015s−1

As we know, hv=hv0+21mv2

i.e.  h(v−v0)=K.E.

∴K.E.=(6.626×10−34)×(1.0×1015−7.0×1014)=6.626×10−34×1014(10−7)

∴K.E=19.878×10−20J

Answer 2

Answer :

Given -

• Threshold frequency = v_0 = 7.0 × 10¹⁴ s^{-1}
• v = Frequency = 1.0 × 10¹⁵ s^{-1}

To Find -

• Kinetic energy ?

Solution -

Einstein's equation for Kinetic energy :

⇒ ½ m_e•v² = h(v - v_0)

(m_e is the mass of electron,

v is the frequency,

h is the Planck's constant,

v_0 is the threshold frequency)

⇒ (6.626 × 10^{-34}) [(1.0 × 10¹⁵) - (7.0 × 10¹⁴)]

⇒ (6.626 × 10^{-34}) [(10.0 × 10¹⁴) - (7.0 × 10¹⁴)]

⇒ (6.626 × 10^{-34}) (3.0 × 10¹⁴)

⇒ 1.988 × 10^{-19} J

Hence, The Kinetic Energy emitted : 1.988 × 10^{-19} J.